Question

1) In a sample of 10 randomly selected individuals, The average hours a person spent smoking cigarettes in a month is 3.6 hours with a standard deviation of 1.2. Determine a 90% confidence interval for the population mean.

2)You recently read an article about a new youth program. This article claimed that at least 45% of high school students encounter some level of bullying on a daily basis. You are shocked at these findings and decide to test this yourself. You sample 200 high school students and find 87 admit to being bullied. Perform a statistical test at the 0.05 level of significance to determine if the article’s claim is true.

Answer 1

Solution :

Given that,

= 3.6

s = 1.2

n = 10

Degrees of freedom = df = n - 1 = 10 - 1 = 9

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t /2,df = t0.05,9 =1.833

Margin of error = E = t/2,df * (s /n)

= 1.833 * (1.2. / 10)

= 0.69

Margin of error = 0.69

The 90% confidence interval estimate of the population mean is,

- E < < + E

3.6 - 0.69 < < 3.6 + 0.69

2.90 < < 4.30

(2.90, 4.30 )

This is the two tailed test .

The null and alternative hypothesis is

H0 : p =0.45

Ha : p 0.45

n = 200

x = 87

= x / n = 87 / 200 =0.435

P0 = 0.45

1 - P0 = 1 - 0.45 =0.55

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

=0.435 - 0.45 / [(0.45*0.55) / 200]

= -0.43

Test statistic = z = -0.43

P(z < -0.43 ) = 0.3336

P-value = 2 *0.3336

= 0.05

P-value >

0.6672 > 0.05

Fail to reject the null hypothesis .

There is insufficient evidence to suggest that