Question

Write solution and answer, please

1) A random sample of 25 junior managers in the offices of corporations in a large city
center was taken to estimate average daily commuting time for all such managers.
Suppose that the population times have a normal distribution with a mean of 87
minutes and a standard deviation of 22 minutes, find the sampling distribution of
sample mean.
3) The student government association at a university wants to estimate the
percentage of the student body that supports a change being considered in the
academic calendar of the university for the next academic year. How many
students should be surveyed if a 90% confidence interval is desired and the margin
of error is to be only 3%?
4) The student government association at a university wants to estimate the
percentage of the student body that supports a change being considered in the
academic calendar of the university for the next academic year. How many
students should be surveyed if a 90% confidence interval is desired and the margin
of error is to be only 3%?
5) A company that receives shipments of batteries tests a random sample of nine of
them before agreeing to take a shipment. The company is concerned that the true
mean lifetime for all batteries in the shipment should be 50 hours. From experience
it is safe to conclude that the population distribution of lifetimes is normal with a
standard deviation of 3 hours. For one shipment the mean lifetime for a sample of
nine batteries was 48.2 hours. Test the hypothesis with = 0.01 level of
significance if the null hypothesis is that the population mean lifetime is 50 hours
6) A wine producer claims that the proportion of its customers who cannot
distinguish its product from frozen grape juice is, at most, 0.09. The producer
decides to test this null hypothesis against the alternative that the true proportion
is more than 0.09. He asked randomly chosen 150 customers and 20 could not
distinguish its product from frozen grape juice. Test the hypothesis with = 0.005
level of significance.
7)The quality-control manager of a chemical company randomly sampled twenty 100-
pound bags of fertilizer to estimate the variance in the pounds of impurities. The sample variance was found to be 6.62. Find a 99% confidence interval for the population variance in the pounds of impurities.
8) An instructor has decided to introduce a greater component of independent study into an intermediate microeconomics course as a way of motivating students to work
independently and think more carefully about the course material. A colleague
cautions that a possible consequence may be increased variability in student
performance. However, the instructor responds that she would expect less variability.
From her records she found that in the past, student scores on the final exam for this
course followed a normal distribution with standard deviation of 18.2 points. For a class of 25 students using the new approach, the standard deviation of scores on the final exam was 15.3 points. If these 25 students can be viewed as a random sample of all those who might be subjected to the new approach, test the null hypothesis that the population standard deviation is at least 18.2 points against the alternative that it islower.
9)

Answer 1

Answer: 1 To find the sampling distribution of the sample mean, we need to use the Central Limit Theorem. According to the Central Limit Theorem, for a large enough sample size, the sampling distribution of the sample mean will be approximately normally distributed, regardless of the shape of the population distribution.

In this case, the population has a normal distribution with a mean of 87 minutes and a standard deviation of 22 minutes. Since the sample size is large (25), we can assume that the sampling distribution of the sample mean will be approximately normally distributed with the same mean but a smaller standard deviation.

To calculate the standard deviation of the sampling distribution (also known as the standard error), we divide the population standard deviation by the square root of the sample size:

Standard Error = Population Standard Deviation / √(Sample Size)

Standard Error = 22 / √(25)

Standard Error = 22 / 5

Standard Error = 4.4 minutes

Therefore, the sampling distribution of the sample mean has a mean of 87 minutes (same as the population mean) and a standard deviation of 4.4 minutes.

3 To determine the sample size needed to achieve a desired margin of error and confidence level, we can use the formula:

Sample Size = (Z-score)^2 * (p * (1 - p)) / (E^2)

In this case, the desired margin of error is 3% (0.03) and the confidence level is 90%, which corresponds to a Z-score of 1.645.

Sample Size = (1.645)^2 * (0.5 * (1 - 0.5)) / (0.03^2)

Sample Size = 2.705 * 0.25 / 0.0009

Sample Size = 6.762 / 0.0009

Sample Size = 7513.33

Therefore, approximately 7514 students should be surveyed to achieve a 90% confidence level with a margin of error of 3%.

5. To test the hypothesis about the population mean lifetime of batteries, we can perform a one-sample t-test.

Null hypothesis (H0): The population mean lifetime of batteries is 50 hours.

Alternative hypothesis (H1): The population mean lifetime of batteries is not 50 hours.

We are given that the sample mean lifetime for a sample of nine batteries was 48.2 hours. The population standard deviation is 3 hours.

Using a significance level of 0.01, we can calculate the t-value and compare it to the critical t-value from the t-distribution with (n-1) degrees of freedom.

t-value = (Sample Mean - Population Mean) / (Sample Standard Deviation / √n)

t-value = (48.2 - 50) / (3 / √9)

t-value = -1.8 / (3 / 3)

t-value = -1.8

With a significance level of 0.01 and (n-1) degrees of freedom (n-1 = 9-1 = 8), the critical t-value is approximately ±2.896.

Since the calculated t-value (-1.8) does not exceed the critical t-value of ±2.896, we fail to reject the null hypothesis. There is not enough evidence to conclude that the population mean lifetime of batteries is significantly different from 50 hours at the 0.01 level of significance.

6.To test the hypothesis about the proportion of customers who cannot distinguish the wine producer's product, we can perform a one-sample proportion test.

Null hypothesis (H0): The proportion of customers who cannot distinguish the product is 0.09 or less.

Alternative hypothesis (H1): The

Explanation: