Question

1.. Give an example of a data set with 6 numbers where the mean is 15 and the median is10.

2. Each of 55 subjects tastes two unmarked cups of coffee and says which he or sheprefers. One cup in each pair contains instant coffee and the other containsfresh-brewed coffee. 35 people prefer fresh-brewed coffee. Is this evidence that amajority of people prefer fresh-brewed coffee?

a.Do a significance test to answer the question with a 10% significance level.

b.Calculate a 95% confidence interval for the percent of coffee drinkers thatprefer fresh brewed coffee.

3. The blood glucose level a person varies throughout the day. Someone with a meanglucose level of more than 200 mg/dL is a diabetic. Diabetes is a serious medicalcondition and may need to be hospitalized or more extensive care.Sally has not been feeling well and her doctor suspects that she is a diabetic and sosends her home with a blood glucose monitor so that she can test her glucose level sover the course of a week. She was instructed to measure her glucose level 40 timesthat week. In so doing she found that her average blood glucose level was...

a.Set up hypotheses for a significance test to determine if Sally is a diabetic.

b.What would a Type I error mean in the context of this problem?

c.What would a Type II error mean in the context of this problem?

If mean= 205.8 mg/dL and the standard deviation of her measurements was s = 10.6mg/dL, what is the value of the test statistic that would be used for a significance test?

Answer 1

1. The following 6 numbers give a mean of 15 and a median of 10.

24,10,9,29,10,8.

2.a. Here the number who preferred freshly brewed coffee x=35

Total number of subjects n=55

The proportion who preferred Fresh coffee

35 p= = 0.6364 n 55

Ho: p=0.5

$H_{1}:p>0.5$(Majority of them preferred freshly brewed)

Level of significance

a = 0.1

The critical region: $R=\left \{ z:z>1.2816 \right \}$

Test statistic: $Z=\frac{\hat{p}-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}$

$Z=\frac{0.6364-0.5}{\sqrt{\frac{0.5(1-0.5)}{55}}}$

  

0.1364 Z = 0.25 55

0.1364 ZE 0.0045

0.1364 Z 0.0674

$Z=2.0231$

Since the calculated value falls in the rejection region, we reject the null hypothesis. Hence, we conclude that there is enough evidence to the claim that "a majority of people prefer fresh-brewed coffee".

(b). 95% confidence interval for the population proportion is

$CI=\hat{p}\mp Z_{1-\alpha/2}*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$CI=0.6364\mp 1.96*\sqrt{\frac{0.6364(1-0.6364)}{55}}$

$CI=0.6364\mp 1.96*\sqrt{\frac{0.6364*0.3636}{55}}$

$CI=0.6364\mp 1.96*\sqrt{\frac{0.2314}{55}}$

$CI=0.6364\mp 1.96*\sqrt{0.0042}$

$CI=0.6364\mp 1.96*0.0649$

$CI=0.6364\mp 0.1271$

$CI=(0.5092,0.7635)$

The 95% confidence interval for the percent of coffee drinkers that prefer fresh brewed coffee is (0.5092,0.7635)

3.

(a). Since Sally was not feeling well, we suspect that she might be diabetic. That is we want to claim whether she is diabetic ie The blood glucose level is more than 200 mg/dL.

$H_{0}:\mu=200$

$H_{1}:\mu>200$(This is the claim)

Test statistic : $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$ will follow a t-distribution with n-1 df.

(b). What would a Type I error mean in the context of this problem?

This means that Sally is diagnosed as diabetic and in fact she is not diabetic.

(c). What would a Type II error mean in the context of this problem.

Sally is not diagnosed as diabetic and actually a diabetic.

Given that $\bar{x}=205.8,s=10.6,n=40$

$t=\frac{205.8-200}{10.6/\sqrt{40}}$

$t=\frac{5.8}{10.6/6.3246}$

$t=\frac{5.8}{1.6760}$

$t=3.4606$

The value of the test statistic is t=3.4606