Question

An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given set of hypotheses and _t_ statistic. Also determine if the null hypothesis would be rejected at $\alpha$

Answer 1

(a) P-value = 0.044

(b) P-value = 0.0022

(c) P-value = 0.4402

(d) P-value = 0.022

Explanation:

The complete question is: An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given set of hypotheses and t-test statistics. Also, determine if the null hypothesis would be rejected at α = 0.05.

a) HA : μ > μ0, n = 11, T = 1.91

b) HA : μ < μ0, n = 17, T = -3.45

c) HA : μ
`\\eq` μ0, n = 7, T = 0.83

d) HA : μ > μ0, n = 28, T = 2.13

(a) We are given the right-tailed test with sample size (n) of 11 and the test statistics of 1.91.

Now, the P-value of the test statistics is given by;

P-value = P(
`t_n_-_1` > 1.91)

= P(
`t_1_0` > 1.91) = 0.044

Since the P-value of the test statistics is less than the level of significance as 0.044 < 0.05, so we have sufficient evidence to reject our null hypothesis.

(b) We are given the left-tailed test with sample size (n) of 17 and the test statistics of -3.45.

Now, the P-value of the test statistics is given by;

P-value = P(
`t_n_-_1` < -3.45)

= P(
`t_1_6` < -3.45) = 0.0022

Since the P-value of the test statistics is less than the level of significance as 0.0022 < 0.05, so we have sufficient evidence to reject our null hypothesis.

(c) We are given the two-tailed test with sample size (n) of 7 and the test statistics of 0.83.

Now, the P-value of the test statistics is given by;

P-value = P(
`t_n_-_1` > 0.83)

= P(
`t_6` > 0.83) = 0.2201

For the two-tailed test, the P-value is calculated as = 2
`*` 0.2201 = 0.4402.

Since the P-value of the test statistics is less than the level of significance as 0.044 < 0.05, so we have sufficient evidence to reject our null hypothesis.

(d) We are given the right-tailed test with sample size (n) of 28 and the test statistics of 2.13.

Now, the P-value of the test statistics is given by;

P-value = P(
`t_n_-_1` > 2.13)

= P(
`t_2_7` > 2.13) = 0.022

Since the P-value of the test statistics is less than the level of significance as 0.022 < 0.05, so we have sufficient evidence to reject our null hypothesis.