A sample of five measurements, randomly selected from a normally distributed population, resulted in the following summary statistics: \(\bar x = 4.8\), \(s = 1.3\) \(\) .

a. Test the null hypothesis that the mean of the population is 6 against the alternative hypothesis, µ<6. Use\(\alpha = .05.\)

b. Test the null hypothesis that the mean of the population is 6 against the alternative hypothesis, µ\( \ne 6\). Use\(\alpha = .05.\)

c. Find the observed significance level for each test.

Referring to pages 412,413 and example 7.5. Given that µ. We must test the null hypothesis.

a)

Test hypotheses are as follows

\(\begin{aligned}\bar x &= 4.8\\s &= 1.3\\n &= 5\\\end{aligned}\)

\(H0:\)µ\( = 6\)

Let’s assume,\(H0::\)µ<\(6\)

Referring to a t distribution table, the test is left tailed. The level of significance is\(0.05\)

Now, the degrees of freedom is\(n - 1\)

\(\begin{aligned}df &= n - 1\\df &= 5 - 1\\df &= 4\end{aligned}\)

Now, the test statistics is

\(\begin{aligned}t &= \frac{{\bar x - \mu }}{{s\sqrt n }}\\t &= \frac{{\left( {4.8 - 6} \right)}}{{2.3\sqrt 5 }}\\t &= - 1.167\end{aligned}\)

The critical value is as follows.

\(\begin{aligned}tc &= T.INV\left( {0.05,4} \right)\\tc &= - 2.132\end{aligned}\)

Rejection region: If t≤\( - 2.132\),reject\(H0\)

So the P-value is above the level of significance, and the null hypothesis cannot be rejected.

Now,the critical value :\(TINV\) \(\left( {0.1,4} \right)\)

The p-value would be:\(TDIST = \left( {1.167,4,1} \right)\)

b)

Test hypotheses are as follows

\(\begin{aligned}H0:\mu &= 6\\Ha:\mu \ne 6\\\alpha &= 0.05\end{aligned}\)

Referring to a t distribution table, the two-tailed test means \(\frac{\alpha }{2}\) because the total\(\alpha \) is 0.05. Keeping the following in mind

The degrees of freedom is as follows.

\(\begin{aligned}df &= n - 1\\df &= 5 - 1\\df &= 4\end{aligned}\)

Now, the test statistics is

\(\begin{aligned}t &= \frac{{\bar x - \mu }}{{s\sqrt n }}\\t &= \frac{{\left( {4.8 - 6} \right)}}{{2.3\sqrt 5 }}\\t &= - 1.167\end{aligned}\)

The right-tailed critical value for\(\frac{\alpha }{2}\) is 2.776 using the t-table.

The rejection region: If\(t \le - 2.776\)or\(t \ge 2.776,\)reject\(H0\)

It cannot be possible to reject the null hypothesis because the test statistics do not fall within the rejection region.

The P-value is: P - Value\( = 0.308\)

So the P-value is greater than the significance level, we cannot reject the null hypothesis.

The critical value\( = TINV\left( {0.05,4} \right)\)

The P-Value\( = TDIST\left( {1.167,4,7} \right)\)

c)

The apparent significance level is also known as the test's p-value.

Part a's observed significance level is 0.1540.

Part a's observed significance level is 0.3080.