Question

According to a report published on a local newspaper in Hong Kong, the average age of viewers of live television programs broadcast on TVB is 51 years old. Suppose a rival network (e.g. ViuTV) executive hypothesizes that the average age of ViuTV viewers is less than 51. To test her hypothesis, she sampled 200 ViuTV viewers and found that their mean age was 50 with a standard deviation of 11.

a. What is the population in this study?

b. What is the variable of interest? Is this variable numerical or categorical?

c. Describe the sample.

d. Describe the inference that the executive of ViuTV is doing in this study.

e. Write down the null and alternative hypotheses for the executive.

f. What is the appropriate probability distribution that should be used in testing the executive’s hypothesis? Give the required condition(s) for using this distribution.

g. At 5% significance level, state the decision rule for the test.

h. Compute the value of the test statistic.

i. Based on the decision rule in part (b) only, what is the statistical decision?

j. Is there sufficient evidence to support the executive’s belief?

Answer 1

Solution :

a) The population in this study consist of all ViuTV viewers.

b) The variable of interest is age of ViuTV viewers. This variable is a numerical variable.

c) Sample consists of age of 200 viewers of viuTV viewers.

d) The executive of viuTV is trying to test the hypothesis that average age of ViuTV viewers is less than 51. The executive is trying to draw the inference about population mean age.

e) The null and alternative hypotheses are as follows :

$\large H_{0}: \mu = 51$ years

$\large H_{1}: \mu < 51$ years

f) Since, the population standard deviation is unknown, therefore the t-distribution should be used in testing the executive's hypothesis.

The conditions required are as follows :

1) The population should be normally distributed.

2) The population standard deviation is unknown.

3) The sample drawn should be a random sample.

g) Significance level = 5% = 0.05

Sample size (n) = 200

Degrees of freedom = (n - 1) = (200 - 1) = 199

Since, our test is left-tailed test, therefore we shall obtain left-tailed critical value of t. The critical value of t at 0.05 significance level and 199 degrees of freedom is given as follows :

$\large Critical Value =- t_{\left(0.05,199\right)} = -1.6525$

For left-tailed test the decision rule would be as follows :

If value of the test statistic is less than -1.6525 (critical value), then we reject the null hypothesis at 5% significance level.

If value of the test statistic is greater than -1.6525 (critical value), then we fail to reject the null hypothesis at 5% significance level.

h) To test the hypothesis we shall use one sample t-test. The test statistic is given as follows :

t = -u s/ vn

Where, x̄ is sample mean, s is sample standard deviation, n is sample size and μ is hypothesized value of population mean under H​​​​​​​0.

We have, n = 200, x̄ = 50, s = 11 and μ = 51

$\large \therefore t = \frac{50-51}{11/\sqrt{200}}=-1.2856$

The value of the test statistic is -1.2856.

i) Decision :

Test statisic value = -1.2856

Critical value = -1.6525

(-1.2856 > -1.6525)

Since, value of the test statistic is greater than -1.6525 (critical value), therefore we shall be fail to reject the null hypothesis (H​​​​​​​​​0) at 5% significance level.

j) Conclusion : At 5% significance level, there is not sufficient evidence to support the executive's claim that the average age of ViuTV viewers is less than 51.

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