Question

Katie works in the film industry and was interested in knowing whether people go to the movies more often in the summer than they do in the winter (assume these data were collected prior to the current pandemic). She collected data from ten people, asking each of them to report how many movies they had gone to the previous summer, and then followed up with them later that year to find out how many movies they went to that winter. Her data are recorded below: Summer: 5,2,5,4,6,3,7,5,5,6 Winter: 5,3,3,4,4,2,4,4,3,4

a.) Conduct the appropriate test of hypothesis, using α = .05.

b.) Generate a 95% confidence interval for these data

c.) if you instead generated a 90% confidence interval, what would happen to the interval? Why?

Answer 1

Answer :

If we take the mean and variance of the summer data we will hav the mean of 4.8 and standard deviation 2.1778

If we take the mean and variance of the winter data we will hav the mean of 3.6 and standard deviation 0.7111

Hypothesis:

Null Hypothesis

People watches the movies in summer is less than or equal to the movies watched in winter

Alternate Hypothesis

People watches the movies in summer is greater than the movies watched in winter

t = mean is summer -mean of winter /sqrt (variance of summer/ sample size of summer+ variance of winter /sampel size of winter)

t = (4.8 -3.6)/sqrt(2.1778/10 + 0.711/10)

t = 1.2/0.5374

t =2.232

so critical t value is 2.232

Answer a:

If we check the critical t value at a =0.05 at 18 degree of freedom, we will have the value of 1.7340 for one tail.

so calculated t value is higher than the critical value so we can reject the null hypothesis and say that movie watched in summer is higher than the winter

Answer b:

as critical t value is 1.7340 so if we have the interval of two standard error on upper side of the mean we will have:

3.6 + 1.7340 **0.5374= 4.53

If the summer movie is less than 4.53, we fail to rejcet the null hypothesis but summer movie is higher than 4.53 so we can accept the alternate hypothesis

Answer C:

If we calcualte the critical value at a =0.1 we have the value of 1.330

So calcauted t value is higher than the critical value so we reject the null hypothesis and accept the alternate hypothesis and say that movie watched in the summer is higer than winter.

As t value at 90% is 1.33 so if we are having 90% of are if we add t value with the winter movie:

upper limit = 3.6 +1.33*0.5374 =4.31

so at 90% we fail to reject the null hypothesis

So if the summer movie is lower than 4.3 we fail to reject the null hypothesis but it is not the case so we can accept the alternate hypothesis