1. The One Sample Z Test

396 people are chosen from a large population that ishalf women. The claim is that the people were randomly chosen, but we suspect that they might not be randomly choosing the people and instead be biased against women. How likely is it that the sample has only 178 women or fewer, if the people were really randomly chosen?

First, How many women would youexpectthe sample to have if it was randomly drawn from a population that is half women?198

You are correct.

Previous Tries

What is the SE for the number of women in the sample?

Tries 0/10

What is the value of the test statistic Z?

Tries 0/10

Now, what is the answer to the question? (What is the p-value?)If you get "Cannot Understand Formula" after submitting an answer, log out of Lon Capa and then log back in. Also, make sure you are using firefox as your browser. If it still doesn't work, take a screenshot of what you're seeing and send us an email.%

Tries 0/10

The One Sample Z Test

A large school district claims that 80% of the children are from low-income families. 130 children from the district are chosen to participate in a community project. Of the 130 only 72% are from low-income families. The children were supposed to be randomly selected. Do you think they really were?

a. The null hypothesis is that the children were randomly chosen. This translates into drawingtimes

Tries 0/3

at random with replacement without replacement

Tries 0/1

from a null box that contains

b.

130 tickets, 72% marked "1" and 28% marked "0"

Thousands of tickets, 80% marked "1" and 20% marked "0"

Thousands of tickets marked either "1" or "0", but the exact percentages of each are unknown and estimated from our sample.

5 tickets, 1 marked "1" and 4 marked "0"

Tries 0/3

c. What is the expected value of the percent of 1's in the draws? (Don't type in the % sign)

%

Tries 0/3

d. What is the SD of the null box?

Tries 0/3

e. What is the standard error of the % of 1's in the draws? (Round to 2 decimal places.)%

Tries 0/3

f. What is the value of the test statistic z? (Round answer to 2 decimal places.)

Tries 0/3

g. What is the p-value?%

Tries 0/3

h. What do you conclude?

There is very strong evidence to reject the null, and conclude that the children were not randomly chosen.

We cannot reject the null, it's plausible the children were randomly chosen.

It is generally claimed that the average body temperature for healthy human adults is 98.6F. To test this claim a simple random sample of 9 adults are chosen. The average temperature of the 9 adults is only 98.1F with a SD of 0.9F. The question is whether the data on these 9 people is sufficient evidence to conclude that average body temperature among all healthy adults is really lower than 98.6. Assume the average body temperatures follow the normal curve.

Compute SD+. (Round to 2 decimal places.)

Tries 0/4

Compute the test statistic. (Round your answer to the nearest tenth, using your previously rounded answer.)

Tries 0/4

How many degrees of freedom are there?

Tries 0/3

Every one is getting different numbers, but most people are getting a p-value between 5-10%. Assume your p-value is about 8%, even though it may not be.

Is the data from the 9 people sufficient evidence to conclude that the average temperature of all healthy adults really lower than 98.6?

Yes,because it means that only 8% of all people have temperatures higher than 98.6.

Yes, because it means that only 8% of all people have temperatures lower than 98.6

No, because 95% of all people have a temperature between 98.6 +/- 8%.

It's impossible to know whether or not to reject the null when the p value is greater than 5%.

A p-value of 8% means that 8% of the time we'll get such a low average or even lower just by the luck of the draw. 8% is not "beyond a reasonable doubt" so we can't reject the null.

Hypothesis Testing

PART 1: Which test-statistic, z, t, or neither?

A candy factory that produces chocolate bars claims each bar weighs 50 grams, at least that is what is printed on the label. Of course, there is bound to be a little variation. An inspector randomly chooses 9 bars from one day's output of 4000 bars. The average weight of the 9 bars is only 47 grams. The inspector wishes to test the null hypothesis that the factory is doing what it is supposed to on this day against the alternative that the company is cheating the consumer. Assume that the weights ofallthe candy bars that are produced follow the normal curve with a SD of 3 gram(s).

Which significance test should be used?

A z-test because the SD of the population is known and the weights follow the normal curve.

You are correct. Your receipt no. is 156-2491

Previous Tries

Do you need to use SD+ to estimate the SD of the population?

No, we're given the SD of the population, we don't need to estimate it.Yes, because with only 9 observations the sample SD is not a great estimate of the true population SD.

Incorrect.

Tries 1/1

Previous Tries

Compute SE. (Give your answer as a decimal to two places)

Tries 0/5

Compute the test statistic. (Round your answer to two decimals, using previously rounded answer.)

Tries 0/5

What do you conclude?

Reject the null, it's very unlikely that the factory is making the candy bars 50 grams as they claim.

Cannot reject the null, it's plausible that the factory is making the bars 50 grams as they claim to be doing.

Tries 0/1

PART 2: Which test-statistic, z, t, or neither?

A candy factory that produces chocolate bars claims each bar weighs 50 grams, at least that is what is printed on the label. Of course, there is bound to be a little variation. An inspector randomly chooses 9 bars from one day's output of 4000 bars. The average weight of the 9 bars is only 47 grams with an SD of 3 gram(s). The inspector wishes to test the null hypothesis that the factory is doing what it is supposed to on this day. You can assume that the weights follow the normal curve.

A z-test because the SD of the population is known and the weights follow the normal curve.A t-test because the SD of the sample is known.A t-test because the SD of the population is unknown, the sample size is less than 25 and the weights follow the normal curve.

Which significance test should be used?

Tries 0/2

Yes, because with only 9 observations the sample SD is not a great estimate of the true population SD.No, we're given the SD of the population, we don't need to estimate it.

Do you need to use SD+ to estimate the SD of the population?

Tries 0/1

Compute SE. (Give your answer as a decimal to two places)

Tries 0/5

Compute the test statistic. (The test statistic is either t or z). Round your answer to 2 decimal places, using previously rounded answers.

Tries 0/5

How many degrees of freedom are there?