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1.For the following pairs, indicate which do not comply with the rules for setting up hypotheses, and explain why. (Select all that apply.) (a) H

1.For the following pairs, indicate which do not comply with the rules for setting up hypotheses, and explain why. (Select all that apply.)

(a)

H₀: = 12, H_a: = 12

This pair complies. This pair does not comply because is not a population characteristic. This pair does not comply because both hypotheses use an equal sign. This pair does not comply because the two hypotheses use different numbers.

(b)

H₀: p = 0.5, H_a: p > 0.7

This pair complies. This pair does not comply because p is not a population characteristic. This pair does not comply because both hypotheses use an equal sign. This pair does not comply because the two hypotheses use different numbers.

(c)

H₀: = 120, H_a: < 120

(d)

H₀: = 127, H_a: = 130

(e)

H₀: p = 0.2, H_a: p 0.2

A certain university has decided to introduce the use of plus and minus with letter grades, as long as there is evidence that more than 60% of the faculty favor the change. A random sample of faculty will be selected, and the resulting data will be used to test the relevant hypotheses. If p represents the proportion of all faculty that favor a change to plus-minus grading, which of the following pairs of hypotheses should the administration test?

H₀: p = 0.6 versus H_a: p > 0.6 H₀: p = 0.6 versus H_a: p < 0.6

Explain your choice.

The university wishes to to determine if less than 60% favor the change. Therefore, the correct alternative is H_a: p > 0.6. The university wishes to to determine if less than 60% favor the change. Therefore, the correct alternative is H_a: p < 0.6. The university wishes to to determine if more than 60% favor the change. Therefore, the correct alternative is H_a: p > 0.6. The university wishes to to determine if more than 60% favor the change. Therefore, the correct alternative is H_a: p < 0.6.

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A researcher speculates that because of differences in diet, Japanese children have a lower mean blood cholesterol level than U.S. children do. Suppose that the mean level for U.S. children is known to be 150. Let represent the true mean blood cholesterol level for Japanese children. What hypotheses should the researcher test?

H₀: = 150 versus H_a: < 160 H₀: = 150 versus H_a: > 150 H₀: = 150 versus H_a: > 140 H₀: = 160 versus H_a: > 150 H₀: = 150 versus H_a: < 150

Pairs of P-values and significance levels, ?, are given. For each pair, state whether you would reject H₀

at the given significance level.

(a)

P-value = 0.064, ? = 0.05

reject fail to reject

(b)

P-value = 0.007, ? = 0.001

reject fail to reject

(c)

P-value = 0.046, ? = 0.05

reject fail to reject

Assuming a random sample from a large population, for which of the following null hypotheses and sample sizes n is the large-sample z test appropriate?

(a)

H₀: p = 0.2, n = 30

appropriate not appropriate

(b)

H₀: p = 0.6, n = 215

appropriate not appropriate

(c)

H₀: p = 0.9, n = 100

appropriate not appropriate

(d)

H₀: p = 0.05, n = 85

appropriate not appropriate

In a survey of 1000 adult Americans, 45.1% indicated that they were somewhat interested or very interested in having web access in their cars. Suppose that the marketing manager of a car manufacturer claims that the 45.1% is based only on a sample and that 45.1% is close to half, so there is no reason to believe that the proportion of all adult Americans who want car web access is less than 0.50. Is the marketing manager correct in his claim? Provide statistical evidence to support your answer. For purposes of this exercise, assume that the sample can be considered as representative of adult Americans. Test the relevant hypotheses using ? = 0.05.

Find the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)

z = P-value =

State the conclusion in the context of the problem.

Reject H₀. We have convincing evidence that the proportion of all adult Americans who want car web access is less than 0.5. The marketing manager is not correct in his claim. Reject H₀. We do not have convincing evidence that the proportion of all adult Americans who want car web access is less than 0.5. The marketing manager is correct in his claim. Fail to reject H₀. We do not have convincing evidence that the proportion of all adult Americans who want car web access is less than 0.5. The marketing manager is correct in his claim. Fail to reject H₀. We have convincing evidence that the proportion of all adult Americans who want car web access is less than 0.5. The marketing manager is not correct in his claim.

You may need to use the appropriate table in the appendix to answer this question.

Plagiarism is a growing concern among college and university faculty members, and many universities are now using software tools to detect student work that is not original.

An Australian university introduced the use of plagiarism detection software in a number of courses. Researchers surveyed 171 students enrolled in those courses. In the survey, 58 of the 171 students indicated that they believed that the use of plagiarism- detection software unfairly targeted students.

Assuming it is reasonable to regard the sample as representative of students at this university, does the sample provide convincing evidence that more than one-third of the students at the university believe that the use of plagiarism- detection software unfairly targets students?

With

p = proportion of all students at the university who believe that the use of plagiarism- detection software unfairly targets students

the relevant hypotheses are

H₀: p =13= 0.33

H_a: p > 0.33

The sample proportion was reported to be p = 0.34.

Does the value of p exceed one-third by enough to cast substantial doubt on H₀?

Because the sample size is large, the statistic

z =p 0.33

has approximately a standard normal distribution when H₀ is true. The calculated value of the test statistic is

z =0.34 0.33=0.010.036= 0.28.

The probability that a z value at least this inconsistent with H₀ would be observed if in fact H₀ is true is

This probability indicates that when p = 0.33,

it would not be all that unusual to observe a sample proportion as large as 0.34. When H₀ is true, roughly 40% of all samples would have a sample proportion larger than 0.34, so a sample proportion of 0.34 is reasonably consistent with the null hypothesis. Although 0.34 is larger than the hypothesized value of p = 0.33,

chance variation from sample to sample is a plausible explanation for what was observed. There is not convincing evidence that the proportion of students who believe that the use of plagiarism- detection software unfairly targets students is greater than one-third.

Suppose that the sample proportion was reported to be p = 0.37.

Calculate a test statistic (z) for the null hypothesis described in the above example.

1.11

1.41

1.15

Which of the following is NOT true?

A P-value is a measure of inconsistency between the hypothesized value for a population characteristic and the observed sample. A P-value is the observed level of significance. A P-value is the probability that the H₀ is incorrect, given the resulting test statistic. A P-value is the probability, assuming that H₀ is true, of obtaining a test statistic value at least as inconsistent with H₀ as what actually resulted.

You may need to use the appropriate table in the appendix or technology to answer this question.

An Associated Press survey was conducted to investigate how people use the nutritional information provided on food package labels. Interviews were conducted with 1,003 randomly selected adult Americans, and each participant was asked a series of questions, including the following two.

Question 1: When purchasing packaged food, how often do you check the nutrition labeling on the package?
Question 2: How often do you purchase foods that are bad for you, even after you've checked the nutrition labels?

It was reported that 582 responded "frequently" to the question about checking labels and 441 responded very often or somewhat often to the question about purchasing "bad" foods even after checking the label.

Let's start by looking at the responses to the first question. Based on these data, is it reasonable to conclude that a majority of adult Americans frequently check the nutritional labels when purchasing packaged foods? We can answer this question by considering

p = actual proportion of adult Americans who frequently check nutritional labels

and testing the following hypotheses.

H₀: p = 0.5

H_a = p > 0.5 (The proportion of adult Americans who frequently check nutritional labels is greater than 0.5. That is, more than half (a majority) frequently check nutritional labels.)

Recall that in a hypothesis test, the null hypothesis is rejected only if there is convincing evidence against itin this case, convincing evidence that p > 0.5. If H₀ is rejected, there is strong support for the claim that a majority of adult Americans frequently check nutritional labels when purchasing packaged foods.

For this sample,

p =5821,003= 0.58.

The observed sample proportion is certainly greater than 0.5, but this could just be due to sampling variability. That is, when p = 0.5

(meaning H₀ is true), the sample proportion p usually differs somewhat from 0.5 simply because of chance variation from one sample to another. Is it plausible that a sample proportion of p = 0.58 occurred as a result of this chance variation, or is it unusual to observe a sample proportion this large when p = 0.5?

To answer this question, we form a test statistic, the quantity used as a basis for making a decision between H₀ and H_a. Creating a test statistic involves replacing p with the hypothesized value in the z variable

z =p p

to obtain

z =p 0.5

If the null hypothesis is true and the sample size is large, this statistic should have approximately a standard normal distribution, because in this case

= 0.5
=(0.5)(1 0.5)n
p has approximately a normal distribution.

The calculated value of z expresses the distance between p and the hypothesized value as a number of standard deviations. If, for example, z = 3, then the value of the sample proportion p is 3 standard deviations (of p) greater than what we would have expected if the null hypothesis were true. How likely is it that a z value at least this inconsistent with H₀ would be observed if in fact H₀ is true? If H₀ is true, the test statistic has (approximately) a standard normal distribution. This means that

P(z 3 when H₀ is true) = area under the z curve to the right of 3.00 = 0.0013.

It follows that if H₀ is true, fewer than 1% of all samples produce a value of z at least as inconsistent with H₀ as z = 3. Because this z value is in the most extreme 1% of the z distribution, it is sensible to reject H₀.

For our data,

z =p 0.5=0.58 0.5=0.080.016= 5.00.

That is, p = 0.58 is 5 standard deviations greater than what we would expect it to be if the null hypothesis H₀: p = 0.5 was true. The sample data appear to be much more consistent with the alternative hypothesis, H_a: p > 0.5. In particular,

P(value of z is at least as contradictory to H₀ as 5.00 when H₀ is true)

= P(z 5.00 when H₀ is true)

= area under the z curve to the right of 5.00

There is virtually no chance of seeing a sample proportion and corresponding z value this extreme as a result of chance variation alone when H₀ is true. If p is 5 standard deviations or more away from 0.5, how can we believe that p = 0.5? The evidence for rejecting H₀ in favor of H_a is very compelling.

Interestingly, in spite of the fact that there is strong evidence that a majority of adult Americans frequently check nutritional labels, the data on responses to the second question suggest that the percentage of people who then ignore the information on the label and purchase "bad" foods anyway is not smallthe sample proportion who responded very often or somewhat often was 0.44.

Suppose that only 540 of those surveyed responded "frequently" to the question about checking labels. Calculate p.

0.74 0.54 0.46 0.58

Suppose that, of those surveyed, only p = 0.53 responded "frequently" to the question about checking labels. Calculate a test statistic z for the null hypothesis described in the above example.

5 1.90 4.25 1.96

You may need to use the appropriate table in the appendix or technology to answer this question.

A report summarized the results of a survey of 309 U.S. businesses. Of these companies, 203 indicated that they monitor employees' web site visits. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of businesses in the United States.

(a)

Is there sufficient evidence to conclude that more than 60% of U.S. businesses monitor employees' web site visits? Test the appropriate hypotheses using a significance level of 0.01.

Find the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)

z= P-value=

State your conclusion.

Fail to reject H₀. We have convincing evidence that more than 60% of U.S. businesses monitor employees' web site visits. Reject H₀. We do not have convincing evidence that more than 60% of U.S. businesses monitor employees' web site visits. Fail to reject H₀. We do not have convincing evidence that more than 60% of U.S. businesses monitor employees' web site visits. Reject H₀. We have convincing evidence that more than 60% of U.S. businesses monitor employees' web site visits.

(b)

Is there sufficient evidence to conclude that a majority of U.S. businesses monitor employees' web site visits? Test the appropriate hypotheses using a significance level of 0.01.

Find the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)

z= P-value=

State your conclusion.

Reject H₀. We do not have convincing evidence that a majority of U.S. businesses monitor employees' web site visits. Fail to reject H₀. We do not have convincing evidence that a majority of U.S. businesses monitor employees' web site visits. Fail to reject H₀. We have convincing evidence that a majority of U.S. businesses monitor employees' web site visits. Reject H₀. We have convincing evidence that a majority of U.S. businesses monitor employees' web site visits.

You may need to use the appropriate table in the appendix to answer this question.

10.

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

Tutorial Exercise

Many consumers pay careful attention to stated nutritional contents on packaged foods when making purchases. It is therefore important that the information on packages be accurate. A random sample of n = 12

frozen dinners of a certain type was selected from production during a particular period, and the calorie content of each one was determined. (This determination entails destroying the product, so a census would certainly not be desirable!) Here are the resulting observations, along with a boxplot and normal probability plot. (Use this dataset for your analysis software.)

255 244 239 242 265 245 259 248 225 226 251 233

The box-and-whisker plot has a vertical axis numbered from 220 to 270. The box-and-whisker is also vertical. The bottom whisker is approximately 225, the bottom edge of the box is approximately 236, the line inside the box is approximately 244.5, the top edge of the box is approximately 253, and the top whisker is approximately 265.

A scatterplot has 12 points. The plot's horizontal axis is labeled "Normal score" and ranges from 1.7 to 1.7. Its vertical axis is labeled "Calories" and ranges from 223 to 267. There is one point plotted near (1.7, 225) and a second near (1.1, 227). After the second point, the 10 remaining points are plotted from left to right in upward, mostly diagonal direction. The pattern begins near the bottom left of the diagram and ends at approximately (1.7, 265). The distance between each point varies slightly. All points are between the approximate horizontal axis values of 1.7 and 1.7 and between the approximate vertical axis values of 223 and 267.

(a)

Is it reasonable to test hypotheses about mean calorie content by using a t test? Explain why or why not.

(b)

The stated calorie content is 239. Does the boxplot suggest that true average content differs from the stated value? Explain your reasoning.

(c)

Carry out a formal test of the hypotheses suggested in part (b).

11.

Many employers are concerned about employees wasting time by surfing the internet and e-mailing friends during work hours. The article "Who Goofs Off 2 Hours a Day? Most Workers, Survey Says" summarized data from a large sample of workers. Suppose that the CEO of a large company wants to determine whether the mean wasted time during an 8-hour work day for employees of her company is less than the mean of 120 minutes reported in the article. Each person in a random sample of 10 employees was contacted and asked about daily wasted time at work (in minutes). Participants would probably have to be guaranteed anonymity to obtain truthful responses. The resulting data are given below.

108 112 117 130 111 131 113 113 105 128 Summary quantities are n = 10, x = 116.80, and s = 9.45.

Do these data provide evidence that the mean wasted time for this company is less than 120 min? To answer this question, let's carry out a hypothesis test with ? = 0.05.

= mean daily wasted time for employees of this company
H₀: = 120
H_a: < 120
? = 0.05
t =x hypothesized values /n=x 120s /n
This test requires a random sample and either a large sample or a normal population distribution. The given sample was a random sample of employees. Because the sample size is small, we must be willing to assume that the population distribution of times is at least approximately normal. The following normal probability plot appears to be reasonably straight, and although the normal probability plot and the boxplot reveal some skewness in the sample, there are no outliers.
A horizontal boxplot has a horizontal axis labeled "Wasted time" with values from 101 to 135 and an unmarked vertical axis. The left whisker is at 105, the left edge of the box is near 111, the line inside the box is approximately 113, the right edge of the box is near 128, and the right whisker is approximately 131.
A scatterplot has 9 points.

The scatter plot's horizontal axis is labeled "Wasted time," and ranges from 100 to 135. Its vertical axis is labeled "Normal score," and ranges from 2.1 to 2.2.
There are 6 points plotted from left to right in upward, mostly diagonal direction. The pattern goes from approximately (105, 1.6) to approximately (117, 0.4). The distance between each point varies.
Three more points are plotted from left to right in upward diagonal direction going from approximately (126, 0.8) to approximately (131, 1.6); the distance between these points varies slightly.
All points are between the approximate horizontal axis values of 104 and 132 and between the approximate vertical axis values of 1.7 and 1.7.

Based on these observations, it is plausible that the population distribution is approximately normal, so we proceed with the t test.
Test statistic:
From the df = 9 column of Table 4 of Appendix A and by rounding the test statistic value to 1.1, we get
P-value = area to the left of 1.1 = area to the right of 1.1 = 0.150
as shown.
A symmetric, mound shaped curve labeled "t curve with df = 9" is graphed above a horizontal axis. The axis has tick marks at 0 and to the left of 0, at 1.1. The curve reaches its maximum above 0. The area under the curve and above the axis is shaded to the left of 1.1 with a label of 0.150.
Because the P-value > ?, we fail to reject H₀. There is not sufficient evidence to conclude that the mean wasted time per eight-hour work day for employees at this company is less than 120 minutes.

Minitab could also have been used to carry out the test, as shown in the output below.

One-Sample T: Wasted Time Test of mu = 120 vs < 120 Variable N Mean StDev SE Mean 95%

Upper

Bound T P Wasted Time 10 116.800 9.449 2.988 122.278 1.07 0.156 Although we had to round the computed t value to 1.1 to use Table 4 of Appendix A, Minitab was able to compute the P-value corresponding to the actual value of the test statistic.

Suppose a random sample was taken from another company and the following output was obtained.

One-Sample T: Wasted Time Test of mu = 120 vs < 120 Variable N Mean StDev SE Mean 95%

Upper

Bound T P Wasted Time 10 114.000 12.157 3.844 122.697 1.56

0.08 State the conclusion of this test.

Because P-value > 0.05 we fail to reject H₀. In this case there is not sufficient evidence to conclude that employees at this company waste less than 120 minutes per 8-hour work day. Because P-value 0.05, we reject H₀. In this case there is not sufficient evidence to conclude that employees at this company waste less than 120 minutes per 8-hour work day. Because P-value > 0.05, we reject H₀. In this case there is sufficient evidence to conclude that employees at this company waste less than 120 minutes per 8-hour work day. Because P-value 0.05, we reject H₀. In this case there is sufficient evidence to conclude that employees at this company waste less than 120 minutes per 8-hour work day.

Suppose we wanted to test the same hypothesis at an significance level of ? = 0.025, what would the conclusion be?

Because P-value > 0.025 we fail to reject H₀. In this case there is not sufficient evidence to conclude that employees at this company waste less than 120 minutes per 8-hour work day. Because P-value 0.025, reject H₀. In this case there is not sufficient evidence to conclude that employees at this company waste less than 120 minutes per 8-hour work day. Because P-value 0.025, we reject H₀. In this case there is sufficient evidence to conclude that employees at this company waste less than 120 minutes per 8-hour work day. Because P-value > 0.025, we reject H₀. In this case there is sufficient evidence to conclude that employees at this company waste less than 120 minutes per 8-hour work day.

You may need to use the appropriate table in the appendix or technology to answer this question.

12.

The authors of the paper "Sex Differences in Time Perception During Smoking Abstinence" carried out a study to investigate how nicotine withdrawal affects time perception and decision-making. In this study, 21 male smokers were asked to abstain from smoking for 24 hours. They were then shown a demo screen with a green cross that changed to a red cross after a period of time. They were then shown the green cross and then asked to indicate when they thought the same amount of time had passed as in the demo. This process was repeated for 15 more trials, with varying times, and then a time discrimination score was calculated as follows for each of the 21 men.

time discrimination score =total estimated timetotal actual time of demos

A time discrimination score greater than 1 would result for someone who tended to overestimate the actual times, and a score of less than 1 would result for someone who tended to underestimate the actual times.

Suppose that the data were as follows (these data are artificial but are consistent with summary quantities given in the paper).

1.12 1.03 1.09 1.03 1.09 0.97 0.98 1.20 1.16 1.03 1.10 1.11 0.98 1.02 1.20 0.96 0.78 1.05 0.90 1.08 0.95 These data were used to calculate the sample mean and standard deviation.

n = 21x = 1.04s = 0.100

Suppose that it is reasonable to consider the people in this sample as representative of male smokers in general. These data can be used to determine if there is evidence that male smokers tend to overestimate time after having abstained from smoking for 24 hours.

With representing the mean time discrimination score for male smokers who have not smoked for 24 hours, we can answer this question by testing

H₀: = 1 (no consistent tendency to overestimate time)

versus

H_a: > 1 (tendency for time to be overestimated).

The null hypothesis is rejected only if there is convincing evidence that > 1. The observed value, 1.04, is larger than 1, but can a sample mean as large as this be plausibly explained by chance variation from one sample to another when = 1?

To answer this question, we carry out a hypothesis test with a significance level of 0.05 using the nine-step procedure described in Section 10.3.

Population characteristic of interest:
= mean time discrimination score for male smokers who have not smoked for 24 hours
Null hypothesis: H₀: = 1
Alternative hypothesis: H_a: > 1
Significance level: ? = 0.05
Test statistic:
t =x hypothesized values/n=x 1s/n
Assumptions: This test requires a random sample and either a large sample size or a normal population distribution. The authors of the paper believed that it was reasonable to consider this sample as representative of male smokers in general, and if this is the case, it is reasonable to regard the sample as if it were a random sample. Because the sample size is only 21, for the t test to be appropriate, we must be willing to assume that the population distribution of time discrimination scores is at least approximately normal. Is this reasonable? The following graph gives a boxplot of the data.
A boxplot has a horizontal axis labeled "Time discrimination score" with values from 0.75 to 1.25. The boxplot is also horizontal. The left whisker is at 0.78, the left edge of the box is approximately 0.98, the line inside the box is at 1.03, the right edge of the box is approximately 1.1, and the right whisker is at 1.2.
Although the boxplot is not perfectly symmetric, it does not appear to be too skewed and there are no outliers, so the use of the t test is reasonable.
Calculations: n = 21, x = 1.04, and s = 0.100, so
t =1.04 10.100/21=0.040.022= 1.82.
P-value: This is an upper-tailed test (the inequality in H_a is "greater than"), so the P-value is the area to the right of the computed t value. Because we can use the df = 20 column of Table 4 in Appendix A to find the P-value. With t = 1.82,
we obtain P-value = area to the right of 1.82. The area to the right of 1.82 is approximately 0.043 (using 1.8, which is the closest value to 1.82 that appears in the table). The P-value 0.043.
Conclusion: Because P-value ?, we reject H₀ at the 0.05 level of significance. It would be unlikely to see a sample mean this extreme as a result of just chance variation when H₀
is true. There is convincing evidence that the mean time discrimination score is greater than 1. This is evidence that male smokers who have not smoked for 24 hours tend to overestimate time.

Suppose the following data was obtained from the sample of nonsmokers:

n = 25x = 0.97s = 0.111

Find the t-statistic associated with the test H₀: = 1, H_a: < 1.

1.32 6.76 1.35 1.35

How would you compute the P-value for this test?

Find the area to the right of the computed t-value with df = 25. Find the area to the right of the computed t-value with df = 24. Find the area to the left of the computed t-value with df = 24. Find the area to the left of the computed t-value with df = 25.

You may need to use the appropriate table in the appendix or technology to answer this question.

13.

Paint used to paint lines on roads must reflect enough light to be clearly visible at night. Let denote the mean reflectometer reading for a new type of paint. A test of H₀: = 20

versus H_a: > 20

based on a sample of 19 observations gave t = 3.1.

What conclusion is appropriate at each of the following significance levels?

(a)

? = 0.05

Reject H₀ Fail to reject H₀

(b)

? = 0.01

Reject H₀ Fail to reject H₀

(c)

? = 0.001

Reject H₀ Fail to reject H₀

You may need to use the appropriate table in the appendix to answer this question.

14.

255 244 239 242 265 245 259 248 225 226 251 233

(a)

Is it reasonable to test hypotheses about mean calorie content by using a t test? Explain why or why not.

Yes, it is reasonable. The pattern in the normal probability plot is roughly linear, and since the sample was a random sample from the population, the t test is appropriate. No, the t test is not applicable here. The pattern in the normal probability plot is roughly linear. Therefore, the t test is not appropriate. Yes, it is reasonable. The pattern in the normal probability plot is not linear, and since the sample was a random sample from the population, the t test is appropriate. No, the t test is not applicable here. The sample was not a random sample from the population so the t test is not appropriate. It depends on the results of t test.

(b)

The stated calorie content is 241. Does the boxplot suggest that true average content differs from the stated value? Explain your reasoning.

Yes, the center of the boxplot is different from the stated value indicating the true average content must differ from the stated value. Not necessarily, it is possible that true average content is 241 and observed differences could be due to sampling variability. No, the true average content must be 241 since 241 is inside the box of the boxplot. Yes, the true average content must be different from 241 since 241 is inside the box of the boxplot.

(c)

Carry out a formal test of the hypotheses suggested in part (b).

Find the test statistic and P-value. (Use ? = 0.05.

Round your test statistic to one decimal place and your P-value to three decimal places.)

t = P-value =

State the conclusion in the context of the problem.

Fail to reject H₀. We do not have convincing evidence that the stated calorie content is not 241. Reject H₀. We have convincing evidence that the stated calorie content is not 241. Fail to reject H₀. We have convincing evidence that the stated calorie content is not 241. Reject H₀. We do not have convincing evidence that the stated calorie content is not 241.

You may need to use the appropriate table in the appendix to answer this question.