The average daily volume of a computer stock in 2011 was muμequals=35.135.1 million shares, according to a reliable source. a stock analyst believes that the stock volume in 2014 is different from the 2011 level. based on a random sample of 4040 trading days in 2014, he finds the sample mean to be 28.628.6 million shares, with a standard deviation of sequals=12.412.4 million shares. test the hypotheses by constructing a 9595% confidence interval. complete parts (a) through (c) below. (a) state the hypotheses for the test.

Required

a. State the hypotheses for the test

b.Construct a 95% confidence interval about the sample mean volume of stocks traded in 2014.

c. Will the researcher reject the null?

Answer

a. Null Hypothesis: μ = 35.1 million

Alternate Hypothesis : μ ≠ 35.1 million

b. CI = 3.708069909 ≤ μ ≤ 53.53193009

c. The critical t value for a two-tailed test at 30 df is ± 2.02. Since the calculated t value at t = -3.315291095 is less than the critical t value, the researcher will reject the null.

b. Since the population standard deviation in unknown, we can construct the confidence interval using t-tables instead of critical Z values.

The formula is:

CI = X̅ ± t₃₉ × (S/√n)

In the term t₃₉, t refers to Student's t table and 39 represents n-1 degrees of freedom (d.f). Since n =40, d.f = 40-1 =39

CI = 28.62 ± 12.7062 × (12.4/√40)

CI = 28.62 ± 24.91193009

CI = 3.708069909 - 53.53193009

c. The test statistic for the test is

t = (X-μ)/ (S/√n)

t = 28.6-35.1/(12.4/√40)

t = -3.315291095

The critical t value for a two-tailed test at 30 df is ± 2.02.

what

explanation:

answer; someone must cover the cost associated with the provision of scarce goods in both sectors;