In a random sample of 150 community college students, the mean number of hours spent studying per week is 11.7 hours and the standard deviation is 4 hours.a) Find the standard score (z-score) for students who study the following hours in a particular week. Round Z to the nearest hundredth and interpret the meaning of each answer as it pertains to this problem.i) 22 hoursii) 6 hoursb)Assuming the distribution of the number of hours community college students study per week is normally distributed, approximately how many of the students in the sample study between 7.7 and 15.7 hours per week?c) Assuming the distribution of the number of hours community college students study per week is normally distributed, approximately how many of the students in the sample study less than 3.7 hours per week?d) Without assuming anything about the distribution of the number of hours community college students study per week, at least what percentage (approximately) of the students study between 5.3 and 18.1 hours per week?e) Without assuming anything about the distribution of the number of hours community college students study per week, at most what percentage of the students study less than 3.7 hours or more than 19.7 hours per week?

In a random sample of 150 community college students, the mean number of hours spentstudying per week is 11.7 hours and the standard deviation is 4 hours.a) Find the standard score (z-score) for…

 

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In a random sample of 25 people, the mean commute time to work was 34.3 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a t-distribution to construct a 99% confidence interval for the population mean. What is the margin of error of the mean? Interpret the results.

The confidence interval for the population mean is ___, ____?

The margin of error of the mean is ____?

 

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In a random sample of ​refrigerators, the mean repair cost was ​$and the population standard deviation is ​$. Construct a ​% confidence interval for the population mean repair cost. Interpret the results. Construct a % confidence interval for the population mean repair cost.

The % confidence interval is ?, ?

Interpret you results. Choose the correct answer below.

A.With % confidence, it can be said that the confidence interval contains the true mean repair cost.

B.The confidence interval contains % of the mean repair costs.

C.With % confidence, it can be said that the confidence interval contains the sample mean repair cost.

 

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In a random sample of 50 ​refrigerators, the mean repair cost was

​$146.00

and the population standard deviation is ​$17.80

.

Construct a 90​% confidence interval for the population mean repair cost. Interpret the results.

Construct a 90% confidence interval for the population mean repair cost.

The 90​% confidence interval is ( ? , ? )

​(Round to two decimal places as​ needed.)

I

nterpret you results. Choose the correct answer below.

A.With 90​% ​confidence, it can be said that the confidence interval contains the true mean repair cost.

B.With 90​% ​confidence, it can be said that the confidence interval contains the sample mean repair cost.

C.The confidence interval contains 90% of the mean repair costs.

 

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Please can you help me with this! I just want to check my answer.

. In a randomized clinical trial to determine the most effective timing of theadministration of chemotherapeutic drugs to lung cancer patients, 26 patients weregiven four drugs simultaneously, while 15 patients were given the same four drugssequentially. Objective response to the treatment (defined as the shrinkage of thetumor by at least 50%) was observed in 16 of the patients treated simultaneously andin 4 of the patients treated sequentially. a. Carry out the appropriate test to determine if there is a difference in effectivenessbetween the two different treatments. Include all steps for full credit. (6pt) b. Quantify the difference in effectiveness of the two treatments with the appropriatemeasure and interpret it in the context of the problem. (2pt)

 

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