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?
<u>Answer</u>
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.
