Question

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1. A researcher wants to prove that counter height in relation to a person's elbow height affects the amount of time it takes to complete a kitchen task. More specifically, the higher the counter is in relation to the person's elbows, the longer it would take to complete a task. A 5% level of significance is used to exam whether or not it takes 10 women of equal height and elbow distance longer to peel a potato with the non-standard 4-foot counter height than with a standard 3-foot counter height. The average time it took the women to peel the potato at the 4-foot counter was 10 seconds while it only took them 5 seconds at the 3-foot counter. The standard deviation of the difference was 7.9 seconds and the standard error was 2.498. Assume an normal distribution. 2. A new facial cleansing system is developed by the New You Beauty Corporation. They claim that the people who use their system regularly and for the same amount of time have the same results as those who use the cleansing system regularly that is manufactured by Hidden Youth Beauty Corporation. A random sample of 12 women who use the New You system show an improvement ranked at 20 with a standard deviation of 4. Of the 10 sampled who use the Hidden Youth system show an improvement ranked at 22 with a standard deviation of 3. Researchers are trying to determine whether the results provided are sufficient evidence to support the claim by New You assuming the population variances are not equal.

Answer 1

1:

Let d = time at 4 foot foot counter - time at 3 foot counter

The average of difference is,

d = 10 - 5 = 5

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Here we have d=5, 5,=7.9, n =10 Hypotheses are: Ho: la 50 H : >0 Level of significance: a =0.05 Test is one tailed (right tailed). Degree of freedom: df =n-1=9 Test Statistics: d-Hd-15-0)/(7.9/5 Ⓡ = (5-0)/(7.9/sqrt(10)) Salin = 2.002 Critical value: Rejection Region: 1.833 Ift > 1.833, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value: P-value=0.0382 Decision: Since p-value is less than level of significance so we reject the null hypothesis Excel function for critical value: Excel function for p-value: ETINV(0.1,9) =TDIST(2.002,9,1)

Conclusion: There is evidence to conclude that it takes longer for women to peel tomato at 4 foot counter.

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(2)

Let level of significance

g= 0,05

Here we need to use independent sample t test. Let sample 1 is sample of new system and sample 2 is sample of Hidden Youth Beauty Corporation.

Following is the test:

Here we have X; =20, S1 =4, n =12 X2=22, 52=3, n=10 Hypotheses are: H:/-1, = 0 H:/4-1 o Test is two tailed. Level of significance: a =0.05 Since we cannot assume that variances are equal so degree of freedom of t test will be: df =- - + =20 (s?/n)" (s22 /n,) n-1 1 -1 Test Statistics: +- (5; -)-(11 – 12) 2) = ((20-22)-0)/(sqrt((16/12)+(9/10))) +- 2 = -1.338 V n n2 Critical value: 2.086 Rejection Region: Ift <= - 2.086 or t >= 2.086 Reject HO Decision: Since test statistics does not lie in rejection region so we fail to reject the null hypothesis. P-value: P-value=0.1959 Decision: Since p-value is not less than level of significance so we fail to reject the null hypothesis. Excel function for critical value: Excel function for p-value: TINV(0.05,20) =TDIST(1.338,20,2)

Conclusion There is sufficient evidence to support the claim by New.