Question

2)

A manufacturer is planning to purchase of a new welder to be used to assemble steel enclosures for products. The size of the weld seam produced by the welder is of interest. The seam widths produced by traditional welding operations have had a mean of 0.15 inch and a standard deviation of 0.010 inch, and the widths are normally distributed. Five trials with the welder produce an average seam width of 0.14 inch. Using a risk level of α = 0.05.

a) What statistic will you use to solve this problem?   
b) What is the value of the statistic? Answer in 2 decimal places.
c) Will you reject the null hypothesis? Answer Yes or No.

3) Dr. Barchard is interested in assessing the Emotional Expressivity of men. Using a sample of 64 men, she found that the mean was 45 and the standard deviation was 16. Calculate a 99% confidence interval for the mean. (use comma to separate the answers, 2 decimal places)

Answer 1

Note : Allowed to solve one question one post.

A manufacturer is planning to purchase of a new welder to be used to assemble steel enclosures for products. The size of the weld seam produced by the welder is of interest. The seam widths produced by traditional welding operations have had a mean of 0.15 inch and a standard deviation of 0.010 inch, and the widths are normally distributed. Five trials with the welder produce an average seam width of 0.14 inch. Using a risk level of α = 0.05.

a) What statistic will you use to solve this problem?
We use t -statistics to solve this problem since the sample size is less than 30

b) What is the value of the statistic? Answer in 2 decimal places.

-2.236

c) Will you reject the null hypothesis? Answer Yes or No.

No.

Since the pvalue is greater than 0.05, we will fail to reject the null hypothesis.

Detailed working

The provided sample mean is I = 0.14 and the sample standard deviation is s = 0.010, and the sample size is n = 5. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: p = 0.15 Ha: u * 0.15 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided the significance level is a = 0.05, and the critical value for a two- tailed test is te = 2.776. The rejection region for this two-tailed test is R={t:t> 2.776} (3) Test Statistics The t-statistic is computed as follows: X - μο. 0.14 -0.15 t= 2.236 0.010/5 (4) Decision about the null hypothesis Since it is observed that t = 2.236 <te = 2.776, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p=0.089, and since p = 0.089 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is different than 0.15, at the 0.05 significance level. 8/ ñ

3) Dr. Barchard is interested in assessing the Emotional Expressivity of men. Using a sample of 64 men, she found that the mean was 45 and the standard deviation was 16. Calculate a 99% confidence interval for the mean. (use comma to separate the answers, 2 decimal places)

99% confidence interval for the mean using T statistics We are given sample mean = pe= 45 sample standard deviation = 0 = 16 n = 64 Confidence interval = 0.99 a/2 = (1-0.99)/2=0.005 df = n - 1 = 63 Using the Student t-table, we find the tstat for a/2 = 0.005 and df = 63 tstat = 2.656145 SE = 16 64 2 16 = 5.3122901 Margin of error = ME = tstat x = 2.656145 Confidence Interval = pi Etstat x = 45+ 2.656145 16 45£ 5.3122901 V 72 Therefore the 99% confidence interval for the mean is (39.6877, 50.3123)