In: Statistics and Probability
Young children in the United States are exposed to an average of 4 hours of background television per day. Having the television on in the background while children are doing other activities may have adverse consequences on a child’s well-being. You have a research hypothesis that children from low-income families are exposed to more than 4 hours of daily background television. In order to test this hypothesis, you have collected a random sample of 64 children from low-income families and found that these children were exposed to a sample mean of 4.5 hours of daily background television. Based on a previous study, you are willing to assume that the population standard deviation is 1.5 hours. Use a .03 level of significance. Calculate and enter the test statistic value. 1e)Type in here the Excel function to be used along with the inputs to the function to calculate p-value for this problem. Calculate in Excel the p-value for this hypothesis test and type in the value here 1f) State the Rejection Criteria under the p-value approach for this problem 1g) What is your decision on the hypothesis test using p-value approach? State the reasoning for your decision. 1h) Type in here the Excel function to be used along with inputs to the function to calculate the Critical-Value for this problem. Calculate in Excel the Critical-Value for this hypothesis test and type in the value here?
Let be the true hours of esposure to background television by the children from low income families.
You have a research hypothesis that children from low-income families are exposed to more than 4 hours of daily background television, that is
The following are the hypotheses that we want to test
We have the following information
n=64 is the sample size of children from low-income families
hours is the sample mean of daily background television.
hours is the population standard deviation
is the standard error of mean
is the hypothesized value of daily background television
Since the sample size is greater than 30 (and we know the population standard deviation) we will use z statistics to test the hypotheses
The sample statistics is
the test statistic value is 2.67
1e) This is a one tailed (right tail) test (the alternative hypothesis has >). The p-value is the area under the right tail and is given by P(Z>2.67) =1-P(Z<2.67).
The excel function used to calculate the p-value is =1-NORM.DIST(2.67,0,1,TRUE)
The p-value from excel is 0.0038
1f). We reject the null hypothesis if the p-value is less than the significance level alpha =0.03.
1g) Since the p-value=0.0038 is less than the significance level, 0.03, we reject the null hypothesis
We conclude that there is sufficient evidence to support the claim that children from low-income families are exposed to more than 4 hours of daily background television.
1h) The level of significance . Since this is a right tail test, we are interested in the value of z for which the area under the right tail is 0.03. That is
The excel function used to calculate this is =NORM.INV(0.97,0,1)
The critical value for this hypothesis test is 1.88
We reject the null hypothesis if the critical value is less than the sample test statistics.
Since the critical value (1.88) is less than the sample statistics of 2.67, we reject the null hypothesis.
That means we conclude that there is sufficient evidence to support the claim that children from low-income families are exposed to more than 4 hours of daily background television.