A researcher performs a hypothesis test to test the claim that for a particular manufacturer, the mean weight of cereal in its 18 ounce boxes is less than 18 ounces. The computer display is shown below. Using a significance level of 0.05, what is your conclusion from these data?

N MEAN STDEV SE MEAN T P VALUE

weight 21 17.883 0.212 0.0463 -2.53 0.01

Question 1 options:

Because the p-value of 0.01 is smaller than the significance level of 0.05, we reject the null hypothesis. There is sufficient evidence to suggest that the mean weight is less than 18 ounces.

Because the p-value of 0.01 is smaller than the signifcance level of 0.05, we reject the null hypothesis. There is sufficient evidence to suggest that the mean weight is equal to 18 ounces.

Because the p-value of 0.01 is smaller than the significance level of 0.05, we fail to reject the null hypothesis. There is insufficient evidence to suggest that the mean weight is less than 18 ounces.

Since x-bar = 17.883, which is clearly less than 18, we should reject the null hypothesis.

Save

Question 2 (1 point) Question 2 Unsaved

Which of the following is NOT part of the necessary condition that must hold in order for the t-distribution to be used for statistical inference?

Question 2 options:

If the sample size is small (n < 15), you can use the t-distribution even if the sample data is skewed.

The data must come from a SRS of the population.

When n is large (n > 40), the t-distribution may be used even when skewness is present in the sample data.

Save

Question 3 (1 point) Question 3 Unsaved

Use the following information to answer the next three questions.

A SRS of 20 recent births at a local hospital were selected. Within the sample, the average birth weight was 121.4 ounces and the standard deviation was 7.5 ounces. Assume that in the population of all babies born at this hospital, the birth weights follow a Normal distribution with a population mean birth weight of mu.

The standard error of the sample mean birth weight is

Question 3 options:

27.1 ounces

6.1 ounces

1.7 ounces

0.4 ounces

Save

Question 4 (1 point) Question 4 Unsaved

After checking to ensure that the sample data follows the necessary condition for using the t-procedure, we use Minitab to get a 90% confidence interval to estimate mu. We find it to be (118.51 ounces, 124.29 ounces). What does this interval tell us?

Question 4 options:

90% of all babies born at this hospital will have birth weights between 118.51 ounces and 124.29 ounces.

We are 90% confident that the average birth weight for the sample of 20 births selected lies between 118.51 ounces and 124.29 ounces.

We are 90% confident that the average birth weight for all babies born at this hospital lies between 118.51 ounces and 124.29 ounces.

There is a 90% chance that a baby born at this hospital will have a birth weight between 118.51 ounces and 124.29 ounces.

Save

Question 5 (1 point) Question 5 Unsaved

If we computed a 95% confidence interval rather than a 90% confidence interval using the same sample of 20 births, how would the 95% confidence interval compare to the 90% one from problem 4?

Question 5 options:

The 95% confidence interval would be wider than the 90% confidence interval.

The 95% confidence interval would be narrower than the 90% confidence interval.

Both of the intervals would be of the same width, but the values for the endpoints of each interval would be different.

There is no way of determining how these two intervals compare without actually recalculating the interval with the data.

Save

Question 6 (1 point) Question 6 Unsaved

Use the following for the next three questions.

A special diet is intended to reduce the cholesterol of patients at risk of heart disease. If the diet is effective, the target is to have the average cholesterol of this group be below 200. After six months on the diet, an SRS of 50 patients at risk for heart disease had an average cholesterol of x-bar = 192, with standard deviation, s = 21. Is this sufficient evidence that the diet is effective in meeting the target? Assume the distribution of the cholesterol for patients in this group is approximately Normal with mean mu.

The appropriate hypotheses for this test are

Question 6 options:

Ho: mu = 200, Ha: mu > 200

Ho: mu = 192, Ha: mu does not equal 192

Ho: mu = 200, Ha: mu does not equal 200

Ho: mu = 200, Ha: mu < 200

Save

Question 7 (1 point) Question 7 Unsaved

The appropriate degrees of freedom for using the t-distribution for this test are

Question 7 options:

21

49

51

200

Save

Question 8 (1 point) Question 8 Unsaved

After conducting this test, the p-value that results is 0.036. Given alpha = 0.05, what conclusion would you make regarding the hypotheses you set up in question 6?

Question 8 options:

Since the p-value is less than the alpha level, we fail to reject the null hypothesis, indicating evidence that the special diet is not effective.

Since the p-value is greater than the alpha level, we fail to reject the null hypothesis, indicating evidence that the special diet is not effective.

Since the p-value is greater than the alpha level, we reject the null hypothesis, indicating evidence that the special diet is effective.

Since the p-value is less than the alpha level, we reject the null hypothesis, indicating evidence that the special diet is effective.

Save

Question 9 (1 point) Question 9 Unsaved

We wish to see if, on average, traffic is moving at the posted speed limit of 65 mph along a certain stretch of Interstate 26. On each of four randomly selected days, a randomly selected car is timed and the speed of the car is recorded. The observed speeds are 70, 65, 70, and 75 mph. Assuming that speeds are normally distributied with mean, mu, we want to test whether the average speed that all cars are traveling is different from 65 mph. We use the following hypotheses for this test:

Ho: mu = 65

Ha: mu does not equal 65

A 95% confidence interval based on the sample is (65.55 mph, 74.45 mph). Based on this interval, what conclusion would you make for this test.

Question 9 options:

Since 65 mph lies outside of the interval, we would conclude to reject the null hypothesis, indicating evidence that the average speed of all cars is not 65 mph, at the 0.05 level.

Since 65 mph lies outside of the interval, we would conclude to fail to reject the null hypothesis, indicating evidence that the average speed of all cars is 65 mph, at the 0.05 level.

Since 65 mph lies inside the interval, we would conclude to fail to reject the null hypothesis, indicating evidence that the average speed of all cars is 65 mph, at the 0.05 level.

Since 65 mph lies inside the interval, we would conclude to reject the null hypothesis, indicating evidence that the average speed of all cars is not equal to 65 mph, at the 0.05 level.

Save

Question 10 (1 point) Question 10 Unsaved

You are going to use the t-distribution to conduct a hypothesis test based on a sample of size 30. You make a histogram of the sample data and find that is shows some moderate skewness, but no outliers are present. Are you safe to continue using the t-distribution with this sample?

Question 10 options:

No, since the sample size is not larger than 40, you should not use the t-distribution.

Yes, since this is considered to be a sample of medium size, using the t-distribution is fine since no outliers are present and only moderate skeweness.

No, for any size sample, if the data is skewed, the t-distribution should not be used.

Save