Instructions: (a) Let alpha be 0.05 for each problem (b) State the null and alte
ID: 3221748 • Letter: I
Question
Instructions: (a) Let alpha be 0.05 for each problem (b) State the null and alternative hypotheses for each problem (c) report the p-value for each problem (d) state the decision for each problem (retain the null or reject the null and accept the alternative).
4. See columns K – M. Is there a significant difference in the mean attention span (in seconds) of 8 – year old children among the three types of TV commercials?
Commercial Type Clothing Food Toys 26 20 60 21 26 51 43 21 43 35 43 54 28 21 63 31 43 53 17 35 48 31 28 58 20 31 47 17 51 31 51 20Explanation / Answer
There a significant difference in the mean attention span (in seconds) of 8 – year old children among the three types of TV commercials to solve this problem we have to use R programming.
alpha = 0.05
(1).Null Hypothesis is
H0: There is a significant difference in the mean attention span of 8 – year old children by a particular TV Programme.
V/S
Alternative Hypothesis is
H1: There is no significant difference in the mean attention span of 8 – year old children by a particular TV Programme.
Now by R programming.
> #input the data
> clothing <- c(26,21,43,35,28,31,17,31,20)
> food <- c(20,26,21,43,21,43,35,28,31,17,31,20)
> Toys <- c(60,51,43,54,63,53,48,58,47,51,51 )
>
> #Start the testing
> # In this problem we have to use the shapiro Test of Normality
> shapiro.test(clothing) # Test for clothing
Shapiro-Wilk normality test
data: clothing
W = 0.96598, p-value = 0.8583
> shapiro.test(food) # Test for food
Shapiro-Wilk normality test
data: food
W = 0.90156, p-value = 0.1661
> shapiro.test(Toys) # Test for Toys
Shapiro-Wilk normality test
data: Toys
W = 0.97054, p-value = 0.892
Conclusion :
In all the cases p-value is greater than the level of significance(alpha)=0.05,so we accept the Null hypothesis.
It means there is a significant difference in the mean attention span of 8 – year old children by a particular TV Programme.