Your local newspaper contains a large number of advertisements for unfurnished one-bedroomapartments in a particular neighborhood. You choose 36 at random and calculate that their meanmonthly rent is $530 with a standard deviation of $78. You want to find out if the sample datagives a good reason to believe that the average rent for all advertised one-bedroom apartments isless than $550 per month. Conduct the following hypothesis test at 0.01 level of significance.Null hypothesis: The population mean rent of the apartments is at least $550 per month.Alternative hypothesis: The population mean rent of the apartments is lower than $550 permonth.None of theseDo not reject the NullAccept the NullReject the Null

The question is about hypothesis testing.Given :Randomly selected no. of apartments ( n ) = 36Sample…

Answered: Your local newspaper contains a large…

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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Justine is an analyst for a sleep study center. She believes that the average adult in a certain city spends more than 7.2 hours sleeping daily. To test this claim, she selects a random sample of 63 adults from the city. The following is the data from this study: The alternative hypothesis Ha:μ>7.2. The sample mean number of hours slept per night by the 63 adults is 7.52 hours. The sample standard deviation is 1.14 hours. The test statistic is calculated as 2.23. Using the information above and the portion of the t− table below, choose the correct p− value and interpretation for this hypothesis test. Values for right-tail areas under the t-distribution curve Probability 0.10 0.05 0.025 0.01 0.005 Degrees of Freedom 60 1.296 1.671 2.000 2.390 2.660 61 1.296 1.670 2.000 2.389 2.659 62 1.295 1.670 1.999 2.388 2.657 63 1.295 1.669 1.998 2.387 2.656 64 1.295 1.669 1.998 2.386 2.655 65 1.295 1.669 1.997 2.385 2.654 66 1.295 1.668 1.997 2.384 2.652…
Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 7 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $16. For 12 randomly selected customers of Company B, you find that they pay a mean of $160 per month with a standard deviation of $14. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.10 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.
Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 15 people who buy insurance from Company A, the mean cost is $154 per month with a standard deviation of $13. For 11 randomly selected customers of Company B, you find that they pay a mean of $159 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.02 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to three decimal places.
Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 12people who buy insurance from Company A, the mean cost is $153 per month with a standard deviation of $16. For 15 randomly selected customers of Company B, you find that they pay a mean of $160 per month with a standard deviation of $10. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.10 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ1=μ2 Ha: μ1_____μ2 Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places Step 3 of…
Utility bills for single family homes are at least $150 per month. After an improvement in infrastructure, the utility company would like to determine if the utility bills are now less than $150 per month. The next 25 utility bills averaged $145.25 per month with the sample standard deviation of $10. At 0.10 significance level, test if the bills are less than $150. 1. Which of the following (A,B,C,or D) would be the correct hypothesis statement? A. Ho: p ≤ 0.25 Ha: p > 0.25 B. Ho: >= 150 H₂: μ 150 D. Ho: M 130 H₁: μ> 130 2. The critical value is 3. The statistical value in this problem is... three decimal places 4. The p-value range associated with the test statistic from the t table is ... 5. At a 10% level of significance, will the decision be to Reject or DNR?
bcd.
In a particular year, registered nurses earned an average annual salary of $52,330. A survey was conducted with 48 nurses from a particular state to determine if the annual salary is higher than $52,330 for that state's nurses. The sample average was $61,599 with a sample standard deviation of $7,489. Conduct a hypothesis test at the 5% level.Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) State the distribution to use for the test. (Enter your answer in the form z or tdf where df is the degrees of freedom.) What is the test statistic? (Round your answer to two decimal places.) What is the p-value? (Round your answer to four decimal places.) Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your answers to the…
Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 1313 people who buy insurance from Company A, the mean cost is $151$⁢151 per month with a standard deviation of $16$⁢16. For 99 randomly selected customers of Company B, you find that they pay a mean of $158$⁢158 per month with a standard deviation of $19$⁢19. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.050.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.
MC Qu. 10-64 Sales at a fast-food restaurant average... Sales at a fast-food restaurant average $6,000 per day. The restaurant decided to introduce an advertising campaign to Increase daily sales. In order to determine the effectiveness of the advertising campaign, a sample of 49 day's sales were taken. The sample showed average daily sales of $6,300. From past history, the restaurant knew that its population standard deviation is about $1,000. If the level of significance is 0.01, have sales increased as a result of the advertising campaign? Multiple Choice Fail to reject the null hypothesis. Reject the null hypothesis and conclude the mean is higher than $6,000 per day. Reject the null hypothesis and conclude the mean is lower than $6,000 per day. Reject the null hypothesis and conclude that mean equal to $6,000 per day.
Toby wants to know the average price of new cars at a particular dealearship. He randomly selects a sample of 31 cars and finds that the average price to be $19500$19500 with a standard deviation of $1640$1640. What is the 95% confidence interval for the mean price of cars at this particular dealership? With 95% confidence, we can say the true population mean price of cars at this dealership is between $____ and _____$ .Round to the nearest dollar
Water usage is closely watched in many parts of Australia. Suppose that in a summer the average household usage in Hobart is 408 litres per day. A year later it examines records of a sample of 50 households and finds that there is a daily mean usage of 380 litres with a standard deviation of 25 litres. a.Do you think the average daily water usage has been changed in the second summer? Use the 5% level of significance level for hypothesis testing. b.Because of the draught situation in Australia, after the first summer, the price of water in Hobart was increased to encourage households to reduce their water consumption. How do you evaluate the effect of increasing water prices? Use the 5% level of significance level for hypothesis testing.

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