When 45 people used a certain diet for one year, their weight losses had a standard deviation of 4.8 Ib. Use a 0.10 significance level to test the claim that the amounts of weight loss have a standard deviation equal to 5.5 Ib, which appears to be the standard deviation for the amounts of weight loss with a different diet. Assume that a simple random sample is selected from a normally distributed population. What are the correct hypotheses for this test? H: o Ib H;: o Ib Calculate the value of the test statistic. O (Round to two decimal places as needed.) Use technology to determine the P-value for the test statistic. The P-value is (Round to three decimal places as needed.) What is the conclusion at the 0.10 level of significance? v the level of significance, V the null hypothesis. There V sufficient evidence to conclude that the sample is from a population with a standard deviation not equal to Since the P-value is V Ib

You are interested in testing whether the average age of household heads is higher in the suburbs than in inner city neighbourhoods. A survey is conducted, and the following results are obtained. In the suburbs, the household heads of the 45 households surveyed had a mean age of 44 with a standard deviation of 15. In the inner city, the mean age of the 40 household heads surveyed was 38 with a standard deviation of 13. Assuming that the populations from which these samples were drawn have equal variances, can you conclude that the difference between the two means is significant at the 95% level of confidence? (d) Calculate the value of t* (e) What is the p-value associated with the test statistic? (f) What conclusion do you reach about the null hypothesis? Explain.

To see if young men ages 8-17 years spend a different amount than the national average of $24.44 per shopping trip to a local mall, the manager surveyed 35 young men. She found that the average amount spent per trip was $23.37 with a standard deviation of $3.70 and alpha = 0.05. What is the most alternative hypothesis?

According to the Center for Disease Control and Prevention (CDC), the mean life expectancy in 2015 for non‑Hispanic black females was 78.1 years. Assume that the standard deviation was 15 years, as suggested by the Bureau of Economic Research. The distribution of age at death, ?, is not normal because it is skewed to the left. Nevertheless, the distribution of the mean, x¯, in all possible samples of size ?n is approximately normal if ? is large enough, by the central limit theorem. Let x¯ be the mean life expectancy in a sample of 100 non‑Hispanic black females. Determine the interval centered at the population mean ? such that 95% of sample means x¯ will fall in the interval. Give your answers precise to one decimal What are the upper and lower limits?

The average expenditure per student (based on average daily attendance) for a certain school year was $10,337 with a population standard deviation of $1560. A survey for the next school year of 150 randomly selected students resulted in a sample mean of $10, 798. Find the Z statistics? Should the null hypothesis be rejected at alpha = .05 level of significance?

Find Test Statistic, P value, and Z score.

Words were displayed on a computer screen with background colors of red and blue. Results from scores on a test of word recall are given below. Use a 0.05 significance level to test the claim that the samples are from populations with the same standard deviation. Assume that both samples are independent simple random samples from populations having normal distributions. Does the background color appear to have an effect on the variation of word recall scores? n X S Red Background 35 15.04 5.97 Blue Background 36 12.93 5.46 What are the null and alternative hypotheses? O A. H₁: 0²₁ = 0²/2 H₁:0²/ ± 0²/2 OB. Ho: 0² #02/2 H₁:0² = 0²/2 O C. H₁: 0²₁2 = 0²/22 OD. H₂:0² = 02²2 H₁ : 0²71 <0²/22 H₁: 021 202²/2 Identify the statistic. F= (Round to two decimal places as needed.) Use technology to identify the P-value. The P-value is. (Round to three decimal places as needed.) What is the conclusion for this hypothesis test? O A. Reject Ho. There is sufficient evidence to warrant rejection of the…

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…

Salaries of 40 college graduates who took a statistics course in college have a​ mean, x overbar, of $64,200. Assuming a standard​ deviation, σ​, of ​$12,042​, construct a 95​% confidence interval for estimating the population mean μ.