Question

3. A consumer watchdog group is studying the number of hours 10-year-olds watch television each day.   A random sample of 25 household containing at least one 10-year-old were surveyed. The parents were asked to record the number of hours the 10-year-old watched television on Saturday, including video games and YouTube.

Mean number of hours watching Television was determined to be 6 and the population standard deviation is known to be 5 hours.

1. What type of data are these? Explain.
2. Can the results be generalized? If so to whom? Explain.
3. Test the hypothesis that the mean number of hours is 8 against the alternate hypothesis that it is not 8. Use an alpha of .05.
4. Develop a 95% confidence interval and interpret in two ways.
5. Determine the p-value and explain its meaning.

Answer 1

All the calculations are done with help of R software and R code and output are given in the last.

a. These data are a primary data because these are collected by consumer watchdog group own self randomly chosen households.

b. The results can not be generalized because here it is not mentioned that from where and which locality these data are collected. And results can be differ from rural area to urban area.

c. Here given hypothesis as H₀: mean number of hours is 8 hours vs H₁: mean number of hours is not equal to 8. So on the basis of given information, we apply one sample t test and we find out that mean number of hours is 8 because value of test statistic lies in the acceptance region. .

d. 95% confidence interval for mean number of hours = (3.936,  8.064), According to this interval we can say that we are 95% confidence that mean number of hours lies between this confidence interval.

e. According to t test, P-value = 0.0569 and level of significance = 0.05, so here P-value is greater than level of significance so we fail to reject null hypothesis is that mean number of hours is 8 hours.

The R code:

####Analysis of data####
n = 25; x_bar = 6; sd_deviation = 5
t_statistic = (((6-8)*5)/5)
t_statistic
t_value = qt(0.025, 24)
t_value
P_value = 2*pt(-2, 24)
P_value
T1_critical = qt(0.975, 24)
T1_critical
T2_critical = qt(0.025, 24)
T2_critical

U = x_bar+T1_critical;U
L = x_bar+T2_critical;L

Output:

####Analysis of data####
> n = 25; x_bar = 6; sd_deviation = 5
> t_statistic = (((6-8)*5)/5)
> t_statistic
[1] -2
> t_value = qt(0.025, 24)
> t_value
[1] -2.063899
> P_value = 2*pt(-2, 24)
> P_value
[1] 0.05693985
> T1_critical = qt(0.975, 24)
> T1_critical
[1] 2.063899
> T2_critical = qt(0.025, 24)
> T2_critical
[1] -2.063899
>
> U = x_bar+T1_critical;U
[1] 8.063899
> L = x_bar+T2_critical;L
[1] 3.936101
>