ALGEBRA 1:
2 answers:
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Answer:
Given:
Mean, u = 135
Sample size, n = 42
Sample mean, x' = 126
Standard deviation = 16
Significance level = 0.05
1) The null and alternative hypotheses:
H0 : u = 135 (the revenue per day is $135)
H1 : u < 135 (the revenue per day is less than $135)
2) In this case, we have a left tailed test.
Let's use the formula:
t = - 3.645
Critical value: at a significance level of 0.05, left tailed test,
tcritical = -1.683
We reject null hypothesis, H0, since t calculated, -3.645 is less than tcritical, -1.683.
3) Given, a = 0.05, df = 41, left tailed test,
t-value = 2.02
For the corresponding confidence interval, we have:
CI = (121.014, 130.986)
4) Conclusion:
We reject the estimated potential revenue advised by the consultant, H0, as it is too high, because the t-calculated falls in rejection region(i.e, less than critical value), also the upper limit of the confidence interval is less than $135.