According to a leasing firm's reports, the mean number of miles driven annually in its leased cars is 13,000 miles with a standard deviation of 3320 miles. The company recently starting using new contracts which require customers to have the cars serviced at their own expense. The company's owner believes the mean number of miles driven annually under the new contracts, μ, is less than 13,000 miles. He takes a random sample of 46 cars under the new contracts. The cars in the sample had a mean of 11,671 annual miles driven. Assume that the population is normally distributed. Is there support for the claim, at the 0.01 level of significance, that the population mean number of miles driven annually by cars under the new contracts, is less than 13,000 miles? Assume that the population standard deviation of miles driven annually was not affected by the change to the contracts.Perform a one-tailed test. Then complete the parts below.Carry your intermediate computations to three or more decimal places, and round your responses as specified below. ### Hypothesis Testing Example for Marriage Counseling EffectivenessA marriage counselor has traditionally seen that the **proportion \( p \)** of all married couples for whom her communication program can prevent divorce is 80%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 80% of married couples. In a **random sample** of 215 married couples who completed her program, 180 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor’s claim at the 0.05 **level of significance**?**Steps to follow:**1. **Perform a one-tailed test**. 2. **Complete the parts below**.Carry your intermediate computations to three or more decimal places. (If necessary, consult a **list of formulas**.)**Graphs or Diagrams:** There are no graphs or diagrams included in this text. This example involves applying statistical hypothesis testing to determine if the marriage counselor's updated program is indeed more effective than the previous one. The relevance of the level of significance helps in making a conclusive decision regarding the hypothesis. ### Finding the Value of the Test StatisticThe value of this test statistic is the z-value corresponding to the sample proportion under the assumption that \( H_0 \) is true. Here is the formula and calculation:\[z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} = \frac{\frac{36}{43} - 0.80}{\sqrt{\frac{0.80(1 - 0.80)}{215}}} \approx 1.364\]In this calculation, \(\hat{p}\) represents the sample proportion, \( p \) is the hypothesized population proportion, and \( n \) is the sample size. The z-value is approximately \( 1.364 \).This calculation involves comparing the difference between the sample proportion and the population proportion, standardized by the standard error of the proportion. The standard error is calculated using the formula \( \sqrt{\frac{p(1-p)}{n}} \).

As per our guidelines we are suppose to answer only one question1) GivenMean(x)=11671standard…

Answered: According to a leasing firm's reports,…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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According to a leasing firm's reports, the mean number of miles driven annually in its leased cars is 13,000 miles with a standard deviation of 3320 miles. The company recently starting using new contracts which require customers to have the cars serviced at their own expense. The company's owner believes the mean number of miles driven annually under the new contracts, μ, is less than 13,000 miles. He takes a random sample of 46 cars under the new contracts. The cars in the sample had a mean of 11,671 annual miles driven. Assume that the population is normally distributed. Is there support for the claim, at the 0.01 level of significance, that the population mean number of miles driven annually by cars under the new contracts, is less than 13,000 miles? Assume that the population standard deviation of miles driven annually was not affected by the change to the contracts.Perform a one-tailed test. Then complete the parts below.Carry your intermediate computations to three or more decimal places, and round your responses as specified below.

### Hypothesis Testing Example for Marriage Counseling Effectiveness

A marriage counselor has traditionally seen that the **proportion \( p \)** of all married couples for whom her communication program can prevent divorce is 80%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 80% of married couples.

In a **random sample** of 215 married couples who completed her program, 180 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor’s claim at the 0.05 **level of significance**?

**Steps to follow:**

1. **Perform a one-tailed test**.
2. **Complete the parts below**.

Carry your intermediate computations to three or more decimal places. (If necessary, consult a **list of formulas**.)

**Graphs or Diagrams:** There are no graphs or diagrams included in this text.

This example involves applying statistical hypothesis testing to determine if the marriage counselor's updated program is indeed more effective than the previous one. The relevance of the level of significance helps in making a conclusive decision regarding the hypothesis.

### Finding the Value of the Test Statistic

The value of this test statistic is the z-value corresponding to the sample proportion under the assumption that \( H_0 \) is true. Here is the formula and calculation:

\[
z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} = \frac{\frac{36}{43} - 0.80}{\sqrt{\frac{0.80(1 - 0.80)}{215}}} \approx 1.364
\]

In this calculation, \(\hat{p}\) represents the sample proportion, \( p \) is the hypothesized population proportion, and \( n \) is the sample size. The z-value is approximately \( 1.364 \).

This calculation involves comparing the difference between the sample proportion and the population proportion, standardized by the standard error of the proportion. The standard error is calculated using the formula \( \sqrt{\frac{p(1-p)}{n}} \).

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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