Question

The article "Plugged In, but Tuned Out"† summarizes data from two surveys of kids age 8 to 18. One survey was conducted in 1999 and the other was conducted in 2009. Data on number of hours per day spent using electronic media that are consistent with summary quantities given in the article are given below (the actual sample sizes for the two surveys were much larger). For purposes of this exercise, assume that it is reasonable to regard the two samples as representative of kids age 8 to 18 in each of the 2 years that the surveys were conducted.

2009 5 9 5 8 7 6 7 9 7 9 6 9 10 9 8

1999 4 5 7 7 5 7 5 6 5 6 7 8 5 6 6

(a) Do the given data provide convincing evidence that the mean number of hours per day spent using electronic media was greater in 2009 than in 1999? Test the relevant hypotheses using a significance level of 0.01. (Use a statistical computer package to calculate the P-value. Use ?2009 ? ?1999. Round your test statistic to two decimal places, your df down to the nearest whole number, and your P-value to three decimal places.)

t =

df =

P-value =

State your conclusion.

Fail to reject H0. There is not convincing evidence that the mean number of hours per day spent using electronic media was greater in 2009 than in 1999. Reject H0. There is convincing evidence that the mean number of hours per day spent using electronic media was greater in 2009 than in 1999. Reject H0. There is not convincing evidence that the mean number of hours per day spent using electronic media was greater in 2009 than in 1999. Fail to reject H0. There is convincing evidence that the mean number of hours per day spent using electronic media was greater in 2009 than in 1999. (b) Construct and interpret a 98% confidence interval estimate of the difference between the mean number of hours per day spent using electronic media in 2009 and 1999. (Use ?2009 ? ?1999. Round your answers to two decimal places.) to hours Interpret the interval. There is a 98% chance that the true mean number of hours per day spent using electronic media in 2009 is directly in the middle of these two values. We are 98% confident that the true mean number of hours per day spent using electronic media in 2009 is between these two values. There is a 98% chance that the true difference in mean number of hours per day spent using electronic media in 2009 and 1999 is directly in the middle of these two values. We are 98% confident that the true mean number of hours per day spent using electronic media in 1999 is between these two values. We are 98% confident that the true difference in mean number of hours per day spent using electronic media in 2009 and 1999 is between these two values.

Answer 1

Please note as nothing is given about population standard deviation, we have performed hypothesis testing and calcualted confidence interval assuming unequal population standard deviation ?₁ ? ?₂. There may be change in calculation if population standard deviation is assumed to be equal.

For doing hypothesis test first we need to find mean and standard deviation of data provided for 2009 and 1999.

For 2009

Mean = Sum of terms/Number of terms

Sum of terms = 5+9+5+8+7+6+7+9+7+9+6+9+10+9+8 = 114

Number of terms n₁ = 15

Mean x?₁ = 114/15 = 7.6000

We will compute standard deviation in 4 steps.

Step 1: Find the mean x?₁

In this case  x?₁ = 7.6000

Standard Deviation s₁ = 1.5946

For 1999

Sum of terms = 4+5+7+7+5+7+5+6+5+6+7+8+5+6+6 = 89

Number of terms n₂ = 15

Mean x?₂ = 89/15 = 5.9333

We will compute standard deviation in 4 steps.

Step 1: Find the mean x?₂

In this case  x?₁ = 5.9333

Standard Deviation s₂ = 1.0998

Hypothesis Testing

As nothing is given about population standard deviation, we have performend hypothesis test assuming unequal population standard deviation ?₁ ? ?₂.

Step 1: Set up null and alternative hypotheses.

H_o: ?₁ = ?₂             (Note: H_o: ?₁ = ?₂ ? H_o: ?₁ - ?₂ = 0)
H_a: ?₁ > ?₂             (Note: H_a: ?₁ > ?₂ ? H_a: ?₁ - ?₂ > 0)

Step 2: Determine ? (level of significance of hypothesis test).

? = 0.01 (Given in question Note: ? = level of significance of hypothesis test = probability of making Type I error.)

Step 3: Calculate test statistic using x?₁, x?₂, s₁, s₂, n₁, and n₂

x?₁ (sample mean 1): 7.6000   
s₁ (sample standard deviation 1) = 1.5946
n₁ (sample size 1) = 15

x?₂ (sample mean 2): 5.9333   
s₂ (sample standard deviation 2) = 1.0998
n₂ (sample size 2) = 15

              (Note: From Step 1, we have H₀: ?₁ = ?₂ ? H₀: ?₁ - ?₂ = 0; therefore, ?₁ - ?₂ = 0)

Test Statistic t = [(7.6000-5.9333) - 0]/SQRT(1.5946²/15+1.0998²/15)

Test Statistic t = 3.3324 = 3.33 (Rounded to 2 decimal places)

Step 4: Determine P-value

Using the test statistic in Step 3 as a t-score, we will find the right-tailed area corresponding to this t-score.
(Note: We use the t-distribution since the population stanadard deviations (?₁ and ?₂) are unknown.)

? = 0.01
Degrees of Freedom = smaller of (n₁ - 1) or (n₂ - 1) = smaller of (15-1 , 15-1) = 14
Right-tailed area corresponding to t-score of 3.3324 = 0.002 (Obtained using p value calculator availabe online for t(14) = 3.3324 screenshot attached)
P-value = Right-Tailed Area = 0.002

Step 5: Make Decision
? = 0.01
P-value = 0.002
In this case, p-value is less than or equal to ?.
If p-value is less than or equal to ? then reject the null hypothesis.

Correct Option is: Reject H₀. There is convincing evidence that the mean number of hours per day spent using electronic media was greater in 2009 than in 1999.

Confidence Interval

As nothing is given about population standard deviation, we have calcualted confidence interval assuming unequal population standard deviation ?₁ ? ?₂.

Step 1: Find ?/2
Level of Confidence = 98%
? = 100% - (Level of Confidence) = 2%
?/2 = 1% = 0.01

Step 2: Find degrees of freedom and t_?/2
Degrees of freedom = smaller of (n₁ - 1 , n₂ - 1 ) = smaller of (14 , 14) = 14
Calculate t_?/2 by using t-distribution with degrees of freedom (DF) = 14 and ?/2 = 0.01 as right-tailed area and left-tailed area.

t_?/2 = 2.624 (Obtained using t distribution table screenshot attached)

Step 3: Calculate Confidence Interval

Standard Error = ? (s?)²/n? + (s?)²/n? = ?0.2501539466666667 = 0.500
Lower Bound = (x?? - x??) - t_?/2•(? (s?)²/n? + (s?)²/n? ) = (7.6 - 5.9333) - (2.624)(0.500) = 0.35
Upper Bound = (x?? + x??) + t_?/2•(? (s?)²/n? + (s?)²/n? ) = (7.6 - 5.9333) + (2.624)(0.500) = 2.98

Confidence Interval = (0.35, 2.98)

Interpretation: We are 98% confident that the true difference in mean number of hours per day spent using electronic media in 2009 and 1999 is between these two values.