Answer to Question #305076 in Statistics and Probability for goawayraine

Question

last year the employees of the city health department donated an average of $ 8 to the rescue squad. test the hypothesis at the 0.01 level of significance that the average contribution this year is still $ 8 if a random sample of 35 employees showed an average donation of $ 8.90 with a standard deviation of $ 1.75.

Answer 1

Hypotheses

H₀: μ= 8                                                 

H_a: μ≠ 8

α = 0.01

Test statistic

Since the sample size n=35 statistically is considered large enough (n >30), the appropriate statistical test is the Z statistic given by the formula:

Z = $\frac{X -\mu}{s/\sqrt{n}}$

Decision rule

From the Z normal distribution table, the critical value for a 99% significance level for two-tailed test is ±2.55. Therefore, we reject the null hypothesis if the computed test statistic Z ≤ -2.55 (lower critical region) or if Z ≥ 2.55 (upper critical region)

Computed test statistic

Given sample mean X̄ = 8.9,

Sample standard deviation s = 1.75 and n=35,

Z = $\frac{X -\mu}{s/\sqrt{n}}$  = $\frac{8.9 -8}{1.75/\sqrt{35}}$  = 3.04

Decision

Since the computed test statistic Z = 3.04>2.55 (the upper critical region), we reject the null hypothesis.

Conclusion

At 0.01 significance level, the results are highly significant suggesting that the data provides sufficient statistical evidence to support the claim that the true population average contribution this year is not equal to $8.