Question

The manager of a paint supply store wants to determine whether the mean amount of paint contained in 1-gallon cans purchased from a nationally known manufacturer is actually 1 gallon. You know from the manufacturer’s specifications that the standard deviation of the amount of paint is 0.02 gallon. You select a random sample of 50 cans, and the mean amount of paint per 1-galllon can is 0.995 gallon.

1. At the .05 level of significance, is there evidence that the mean amount is different from 1.0 gallon? Show all steps! Draw the normal curve and insert all information given.
2. Compute the p-value and interpret its meaning. Show all steps!
3. Construct a 95% confidence interval estimate of the population mean amount of paint. Show all steps!
4. Compare the results of (a) and (c). What conclusions do you reach? Show all steps and graphs
   

Answer 1

a) - b)

c)

d) Since the confidence interval includes 1, we fail to reject H₀ at 95% level of confidence.

/one sample_t shiny Shiny http://127.0.0.1:7521 Open in Browser Let u denotes the population mean. To test null hypothesis Ho 1.0 against alternative hypothesis H1 1.0 Let denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, 0.9950, s20.0004, s 0.0200, n = 50 The test statistic can be written as Vn( 1.0) S which under Ho follows a t distribution with n-1 df. We reject Ho at 0.050 level of significance if P-value 0.050 or if |tobs > to.025,n-1 Now The value of the test statistic= -1.767767 associated degrees of freedom 49 The critical value = 2.009575 value P(t,-1> tohs)= 2 * P(t49 < -1.767767) = 2 * 0.041663 0.083326 Since p-value > 0.050 and tobsteritical = 2.009575, so We fail to reject Ho at 0.050 level of significance Hence, we can conclude that the population mean is not different from 1.0

denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size Let Here, 0.995, s = 0.0004, s = 0.0200, n = 50 a 95 confidence interval for the population mean is = x ± Vn t1-0,950 49 2 0.020 * to.025,49 0.995 50 0.020 0.995 2.009575 50 0.995 ± 0.005684 (0.989316, 1.000684)