Question

19 25 The sample of six measurements shown below was randomly selected from a normally distributed population. Complete parts a through c. 1,2,3,3,4,1 a. Test the null hypothesis that the mean of the population is 3 against the alternative hypothesis. p < 3. Use a = 0.05 Ifq=0.05, find the rejection region for the test. Choose the correct answer below. % 1994 1994 OA. <-2015 or t> 2015 Oct-2571 O E. > 2571 OB < -2015 OD < -2571 ort > 2.571 OF > 2.015 Calculate the value of the test statistic 1- (Round to two decimal places as needed) Make the appropriate conclusion. Choose the correct answer below. O A Reject Hy. There is insufficient evidence at the a= 0.05 level of significance to conclude that the true mean of the population is less than 3 Click to select your answer(s) Relal

Answer 1

X follows the normal distribution.

Sample (xi) and statistics are given by -

xi	X; - 79
1 2 3 3 4 1	1.7689 0.1089 0.4489 0.4489 2.7889 1.7689
Ti = 14	$\sum (x_i - \bar{x})^{2}$= 7.3334

Sample mean is -

$\bar{x}=\frac{\sum x_i}{n} = \frac{14}{6}=2.33$

Sample standard deviation is -

​​​​​​

Στη - )2 η – 1 7.3334 5 V1.467 = 1.211

Sample size = n = 6

Degrees of freedom = n - 1 = 5

a) Hypothesis can be framed as -

Null hypothesis (Ho) : $\mu =3$

Alternative hypothesis (H1) : $\mu <3$

Test statistic is given by -

$t =\frac{\bar{x}-\mu}{s/\sqrt{n}}$

where, $\mu$ is the specific value of population mean under the null hypothesis = 3

Hence, the value of test statistic will be -

$t = \frac{2.33-3}{1.211/\sqrt{6}}$

  $=\frac{-0.666}{0.494}$

= -1.35

So, the value of test statistic, t = -1.35

Since, level of significance = $\alpha$ = 0.05

The critical value of t with 5 degrees of freedom at 0.05 level of significance for the lower tailed test is 2.015 (as obtained from the t table corresponding to 5 degrees of freedom and 0.05 probability)

So, the rejection region will be t < -2.015 (Since, it is lower tailed test or left tailed test).

Hence, option (B) is the correct option.

Also, since, our test statistic > critical value of t, it doesn't lie in the rejection region, hence, null hypothesis may not be rejected, hence, true population mean,  $\mu =3$

So, conclusion will be : Do not reject Ho, there is insufficient evidence at $\alpha = 0.05$ level of significance to conclude that the true mean of population is less than 3.

Hence, option (c) is correct.

b) To test :

Ho : $\mu =3$

H1 : $\mu \neq 3$

The critical value of t with 5 degrees of freedom at 0.05 level of significance for the two tailed test is 2.571 (as obtained from the t table corresponding to 5 degrees of freedom and 0.025 probability, as it is two tailed so 0.025 probability will be on both sides)

So, the rejection region will be -

t < -2.571 or t > 2.571

Hence, option (f) is the correct option.

Test statistic will be -

$t =\frac{\bar{x}-\mu}{s/\sqrt{n}}$

$= \frac{2.33-3}{1.211/\sqrt{6}}$

= -1.35 that is same as calculated in part (a)

Since, the value of the test statistic does not lie in the rejection region, hence, the null hypothesis may not be rejected, so, the true population mean is 3.

So, our conclusion is :

Do not reject Ho, there is insufficient evidence at $\alpha = 0.05$ level of significance to conclude that the true mean of the population is not 3.

Hence, option (c) is the correct option.

c) P value for part (a) is given by -

P value = P(t $\leq$ -1.35) = the area under the t curve with 5 degrees of freedom on the left side of t = -1.35.

This can be obtained from the t table by finding the probability at 5 degrees of freedom for which t = 1.35 and will lie somewhere between 0.10 and 0.15 say 0.12

P value = 0.12

Now, P value for part (b) will be -

P(|t| $\geq$ 1.35) = P(t $\leq$ -1.35) + P(t $\geq$ 1.35)

That is the area under the t curve with 5 degrees of freedom on the left side of t = -1.35 and right side of t = 1.35.

[Now, P(t $\leq$ -1.35) = P(t $\geq$ 1.35) as t curve is symmetric about the mean , and, we have calculated P(t $\leq$ -1.35) for p value of part (a) as 0.12]

So, P value = $P(t\leq -1.35) + P (t \geq 1.35) = 0.12 + 0.12= 0.24$