Question

A sample of 148 college students at a large university reports getting an average of 6.85 hours of sleep last night with a standard deviation of 2.12 hours.
a.Verify that it is reasonable to use the t-distribution to construct a confidence interval for the average amount of sleep students at this university got last night.
b. Construct a 98% confidence interval for the average amount of sleep students at this university got last night. Use two decimal places in your margin of error.
c.. Provide an interpretation of your interval in the context of this data situation.
d.. Suppose you want to conduct a similar study at your university. Assuming that the standard deviation of this sample is a reasonable estimate of the standard deviation of sleep time at your university, how many students do you need to survey to estimate the mean sleep time of students at your university with 95% confidence and a margin of error of 0.5 hours?

Answer 1

The solution for the questions is mathematically given as

a)

t-distribution.

b)

the confidence interval for the mean, based on 98 percent of the sample, is ( 6.3952, 7.3048 )

c) the value of the $\mu_0$ is within the range of the 98 percent confidence interval for the mean, which is between 6.3952 and 7.3048, then accept H_0; otherwise, reject H _0.

d)

you should conduct a poll with around 123 students to determine the average amount of time that students spend sleeping at your institution.

What is the distribution to use?

Generally, the equation for is mathematically given as

a.

In this case, the standard deviation of the population is unknown.

As a result, we make use of the t-distribution.

b)

We wish to generate a confidence interval with a 98 percent likelihood for the mean.

Because of this,

$(\bar{X}-t_{n-1,\alpha/2}\frac{s}{\sqrt{n}},\bar{X}+t_{n-1,\alpha/2}\frac{s}{\sqrt{n}})$

$(6.85-t_{148-1,0.02/2}\frac{2.12}{\sqrt{148}},6.85+t_{148-1,0.02/2}\frac{2.12}{\sqrt{148}})$

$(6.85-t_{147,0.01}\frac{2.12}{12.1655},6.85+t_{147,0.01}\frac{2.12}{12.1655})$

(6.3952,7.3048)

Therefore, the confidence interval for the mean, based on 98 percent of the sample, is ( 6.3952 , 7.3048 )

c )

If the value of the mu _0 is within the range of the 98 percent confidence interval for the mean, which is between 6.3952 and 7.3048, then accept H_o; otherwise, reject H_0.

d . Here, we want to determine the sample size

Therefore,

$n=t_{n-1,\alpha/2}^2\frac{s^2}{E^2}$

$n=t_{148-1,0.05/2}^2\frac{2.12^2}{0.5^2}$

$n=t_{147,0.025}^2\frac{2.12^2}{0.5^2}$

$n=2.6097^2\frac{2.12^2}{0.5^2}$

n=122.4364

In conclusion, you should conduct a poll with around 123 students to determine the average amount of time that students spend sleeping at your institution.

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