Question

In a test of the hypothesis \({H_0}:\mu = 10\) versus \({H_a}:\mu \ne 10\), a sample of n = 50 observations possessed mean \(\bar x = 10.7\) and standard deviation s = 3.1. Find and interpret the p-value for this test.

Short answer

The p-value for the two-tailed test is 0.1096. When the null hypothesis \(\mu = 10\) is true, the probability that the test statistic is more than 1.60 is 0.1096.

Step-by-step solution

Given information

The hypothesis test is:\({H_0}:\mu = 10\)versus\({H_a}:\mu \ne 10\).

A random sample of size 50 is selected.

The sample mean is\(\bar x = 10.7\).

The sample standard deviation is s=3.1.

Computing the value of the test statistic

The test statistic is:

\(\begin{aligned}z &= \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\\ &= \frac{{10.7 - 10}}{{\frac{{3.1}}{{\sqrt {50} }}}}\\ &= \frac{{0.7}}{{0.4384}}\\ &= 1.5967\\ &= 1.60\end{aligned}\).

The test statistic is \(z = 1.60\).

Computing the p-value

The p-value for the two-tailed test is:

\(\begin{aligned}p &= 2P\left( {Z > 1.60} \right)\\ &= 2\left( {1 - P\left( {Z \le 1.60} \right)} \right)\\ &= 2 \times \left( {1 - 0.9452} \right)\\ &= 2 \times 0.0548\\ &= 0.1096\end{aligned}\).

The z-table is used to obtain the probability of a z-score less than or equal to 1.60.

The p-value for the two-tailed test is 0.1096.

Interpretation of the p-value

When the null hypothesis \(\mu = 10\) is true, the probability that the test statistic is more than 1.60 is 0.1096.