Question

A pressure-relief valve (PRV) is installed on a residential hot-water heater to protect against overheating and of course, ligh pressure. Independent random samples of PRVs from different companies were obtained. Each value was tested by recording the pressure (psi) required to cause the valve to open. 1 statistics are shown below. he summary Sample SizeSample Msan Company Delta Gamma 28 147.6 3.09 25 148.8 16.70 Assumin g normality, test at the 5% significance level to determine if there is evidence to suggest that the mean pressure required to open valves produced by Delta is lower than that of Gamma (18 points) Assumptiots Test Statistie Rejection Region Conclusion P-value If you had tested at the 10% significance leveL would your conclusion have changed? Explain

Answer 1

it is assumed that the population variances are equal

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ < μ2​

This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$t = \frac{\bar X_1 - \bar X_2}{\sqrt{ \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}(\frac{1}{n_1}+\frac{1}{n_2}) } }$

$= \frac{ 147.6 - 148.8}{\sqrt{ \frac{(28-1)3.09 + (25-1)16.70}{ 28+25-2}(\frac{1}{ 28}+\frac{1}{ 25}) } } = -1.415$

DF=n1+n2-2=51

critical value for this left-tailed test is

tc​=−1.675, for α=0.05 and df = 51

The rejection region for this left-tailed test isR={t:t<−1.675}

p-value=0.0815

Decision about the null hypothesis

Since it is observed that t=−1.415≥tc​=−1.675, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.0815, and since p=0.0815≥0.05, it is concluded that the null hypothesis is not rejected

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at alpha=0.10

p-value=0.0815<0.10,so null hypothesis rejected.