Question

A group of 43 college students from a certain liberal arts college were randomly sampled and asked about the number of alcoholic drinks they have in a typical week. The purpose of this study was to compare the drinking habits of the students at the college to the drinking habits of college students in general. In particular, the dean of students, who initiated this study, would like to check whether the mean number of alcoholic drinks that students at his college in a typical week differs from the mean of U.S. college students in general, which is estimated to be 4.73.

The group of 43 students in the study reported an average of 5.52 drinks per with a standard deviation of 3.64 drinks.

Find the p-value for the hypothesis test.

The p-value should be rounded to 4-decimal places.

Commute times in the U.S. are heavily skewed to the right. We select a random sample of 220 people from the 2000 U.S. Census who reported a non-zero commute time.

In this sample the mean commute time is 28.3 minutes with a standard deviation of 19.2 minutes. Can we conclude from this data that the mean commute time in the U.S. is less than half an hour? Conduct a hypothesis test at the 5% level of significance.

What is the p-value for this hypothesis test?

Your answer should be rounded to 4 decimal places.

Dean Halverson recently read that full-time college students study 20 hours each week. She decides to do a study at her university to see if there is evidence to show that this is not true at her university. A random sample of 33 students were asked to keep a diary of their activities over a period of several weeks. It was found that the average number of hours that the 33 students studied each week was 22.2 hours. The sample standard deviation of 3.9 hours.

Find the p-value.

The p-value should be rounded to 4-decimal places.

A medical researcher is studying the effects of a drug on blood pressure. Subjects in the study have their blood pressure taken at the beginning of the study. After being on the medication for 4 weeks, their blood pressure is taken again. The change in blood pressure is recorded and used in doing the hypothesis test.

Change: Final Blood Pressure - Initial Blood Pressure

The researcher wants to know if there is evidence that the drug affects blood pressure. At the end of 4 weeks, 33 subjects in the study had an average change in blood pressure of -2.9 with a standard deviation of 5.4.

Find the p-value for the hypothesis test.

Your answer should be rounded to 4 decimal places.

Find the p-value for the hypothesis test. A random sample of size 53 is taken. The sample has a mean of 369 and a standard deviation of 84.

H0: µ = 400

Ha: µ< 400

The p-value for the hypothesis test is .

Your answer should be rounded to 4 decimal places.

Child Health and Development Studies (CHDS) has been collecting data about expectant mothers in Oakland, CA since 1959. One of the measurements taken by CHDS is the weight increase (in pounds) for expectant mothers in the second trimester.

In a fictitious study, suppose that CHDS finds the average weight increase in the second trimester is 14 pounds. Suppose also that, in 2015, a random sample of 37 expectant mothers have mean weight increase of 16.2 pounds in the second trimester, with a standard deviation of 5.6 pounds.

A hypothesis test is done to see if there is evidence that weight increase in the second trimester is greater than 14 pounds.

Find the p-value for the hypothesis test.

The p-value should be rounded to 4 decimal places.

Answer 1

Let denote the average number of drinks per week by College students.

Vs

Given:

To test the above hypothesis, the appropriate statistical test would be a One sample t test.Assuming all the assumptions of test is satisfied.

The test statistic can be computed as:

= 3.387

To obtain the p-value, using excel,

we get, p- value = 0.0015<0.05.

We may conclude that mean number of alcoholic drinks consumed by students at college in a typical week differs from 4.73.

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Let denote the mean commute time.

Vs

Given:

To test the above hypothesis, the appropriate statistical test would be a One sample t test.Assuming all the assumptions of test is satisfied.**

The test statistic can be computed as:

= -1.313

To obtain the p-value, using excel, (used 1.313 instead of (-1.313) in excel for computational ease; since, p-value would be the same for both since t distribution is symmetric)

we get, p- value = 0.0953 > 0.05

We may conclude that we do not have sufficient evidence to conclude that the mean commute time in the U.S is less than half hour.

** However, it is stated in the start of the problem that the population data is highly skewed to the right, hence, the test would not be reliable, here; we might go for its non parametric alternative Wilcoxon signed rank test. We may run the test, given the whole data set, since its test statistic dependes on the rank of the observations.

But here, since the data is not given, we have computed the p-value for one sample t test.

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Let denote the average number of study hours

Vs

Given:

To test the above hypothesis, the appropriate statistical test would be a One sample t test.Assuming all the assumptions of test is satisfied.

The test statistic can be computed as:

= 3.241

To obtain the p-value, using excel,

we get, p- value = 0.0028<0.05.

We may conclude that mean study hours is different from 20 hours.

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Let denote the mean change in blood pressure

Vs

Given:

To test the above hypothesis, the appropriate statistical test would be a paired t test.Assuming all the assumptions of test is satisfied.

The test statistic can be computed as:

= -3.085

To obtain the p-value, using excel,

we get, p- value = 0.0042<0.05.

We have enough evidence to conclude that the drug significantly affects blood pressure.

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Let denote the population mean.

Vs

Given:

To test the above hypothesis, the appropriate statistical test would be a One sample t test.Assuming all the assumptions of test is satisfied.

The test statistic can be computed as:

= -2.687

To obtain the p-value, using excel,

we get, p- value = 0.0048<0.05

We have enough evidence to conclude that the mean is less than 400.

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Let denote the average weight increase of expectant mothers in the second trimester

Vs

Given:

To test the above hypothesis, the appropriate statistical test would be a paired t test.Assuming all the assumptions of test is satisfied.

The test statistic can be computed as:

= 2.390

To obtain the p-value, using excel,

we get, p- value = 0.0111<0.05.

We may conclude that there is enough evidence that weight increase in the second trimester is greater than 14 pounds.