Question

Jody is interested in the population mean of pets per household in her neighborhood. She hypothesized that the average amount of pets per household will be less than 3.

To test the hypothesis, she used the following data.

0 1 3 2 1 3

1 0 4 1 0 5

1 2 0 1 2 1

Jody's hypotheses are

H₀ : µ≥3

H_A : µ<3

Please calculate 1) null and alternative hypothesis 2) alpha value 3) p-value 4) conclusion

Answer 1

Solution:-

	Pets
Sum	28
Mean	1.55555555555556
S.D	1.4234267774809
Count	18
S.E	0.335504908959753

1) State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u > 3
Alternative hypothesis: u < 3

Note that these hypotheses constitute a one-tailed test.

2) Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.3355
DF = n - 1

D.F = 17
t = (x - u) / SE

t = - 4.31

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

3) The observed sample mean produced a t statistic test statistic of - 4.31.

Thus the P-value in this analysis is 0.00024.

Interpret results. Since the P-value (0.00024) is less than the significance level (0.05), hence we have to reject the null hypothesis.

4) From the above test we have sufficient evidence in the favor of the claim that the average amount of pets per household will be less than 3.