Question

1) A new cream that advertises that it can reduce wrinkles and improve skin was subject to a recent study. A sample of 63women over the age of 50 used the new cream for 6 months. Of those 63women, 36 of them reported skin improvement(as judged by a dermatologist). Is this evidence that the cream will improve the skin of more than 60% of women over the age of 50? Test using ?=0.01

(a) Test statistic: ?=

(b) Critical Value: ?∗=

2) A survey of 1325 people who took trips revealed that 140 of them included a visit to a theme park. Based on those survery results, a management consultant claims that less than 12% of trips include a theme park visit. Test this claim using the ?=0.01 significance level.

The test statistic is ?

The critical value is?

3) A new cream that advertises that it can reduce wrinkles and improve skin was subject to a recent study. A sample of 54 women over the age of 50 used the new cream for 6 months. Of those 54 women, 32 of them reported skin improvement(as judged by a dermatologist). Is this evidence that the cream will improve the skin of more than 60% of women over the age of 50? Test using ?=0.05

test statistics ?=

rejection region ?>

Answer 1

Solution :

1) Given that,

n = 63

x = 36

The null and alternative hypothesis is

H₀ : p = 0.60

H_a : p > 0.60

This is right tailed test.

= x / n = 36 / 63 = 0.571

P₀ = 60% = 0.60

1 - P₀ = 1 - 0.60 = 0.40

(a) Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.571 - 0.60 / [(0.60 * 0.40) / 63]

= -0.463

Test statistic : z = -0.463

(b) = 0.01

z  = z_0.01  = 2.33

Critical value : z = 2.33

2) Given that,

n = 1325

x = 140

The null and alternative hypothesis is

H₀ : p = 0.12

H_a : p < 0.12

This is left tailed test.

= x / n = 140 / 1325 = 0.1057

P₀ = 12% = 0.12

1 - P₀ = 1 - 0.12 = 0.88

(a) Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.1057 - 0.12 / [(0.12 * 0.88) / 1325]

= -1.606

Test statistic is -1.606

(b) = 0.01

z  = z_0.01  = -2.33

Critical value : z = -2.33

3) Given that,

n = 54

x = 32

The null and alternative hypothesis is

H₀ : p = 0.60

H_a : p > 0.60

This is right tailed test.

= x / n = 32 / 54 = 0.5926

P₀ = 60% = 0.60

1 - P₀ = 1 - 0.60 = 0.40

(a) Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.5926 - 0.60 / [(0.60 * 0.40) / 54 ]

= -0.111

Test statistic z is -0.111

(b) = 0.05

z  = z_0.05  = 1.64

Critical value : z = 1.64

The rejection region for this right-tailed test is = { z > 1.64 }