ass ociated levels of confidence, ct to compute probabilities from computed z or

Question

ass ociated levels of confidence, ct to compute probabilities from computed z or t values, when applicable.... #1) A report by the Gallup Poll a few years ago stated that, on average, a woman visits her physician 5.8 times per year. A local researcher randomly selects women in Minneapolis and obtained this data for the number of visits to a physician in a year: Using the sample data, construct a 95 % confidence interval for the true mean number of times a woman in Minneapolis visits her physician in a year. How does your answer compare to the Gallup Poll results ? Explain fully. Show all work. Be complete

Explanation / Answer

Question 1

Here, we have to find 95% confidence interval for true mean.

Confidence interval = Xbar -/+ t*S/sqrt(n)

From given data, we have

Xbar = 3.7

S = 2.556724879

n = 20

c = 0.95

= 1 – c = 1 – 0.95 = 0.05

df = n – 1 = 20 – 1 = 19

Critical value = t/2, n – 1 = 2.0930

Confidence interval = 3.7 -/+ 2.0930*2.556724879/sqrt(20)

Confidence interval = 3.7 -/+ 1.1966

Lower limit = 3.7 - 1.1966 = 2.50

Upper limit =3.7 + 1.1966 = 4.90

The average number of visits for Gallup poll is given as 5.8 which is not included in the above confidence interval. So, we reject the null hypothesis that average number of visits is 5.8.

Question 2

Part a

We are given

X = 24

n = 250

P = X/n = 24/250 = 0.096

= 0.06

Here, we have to use z test for population proportion.

H0: p = 0.05 versus Ha: p 0.05

This is a two tailed test.

Test statistic = Z = (P – p) / sqrt(pq/n)

Where, q = 1 – p = 1 – 0.05 = 0.95

Test statistic = Z = (0.096 – 0.05) / sqrt(0.05*0.95/250)

Z = 3.3372

Lower critical value = -1.8808

Upper critical value = 1.8808

Test statistic Z is out of critical values, so we reject the null hypothesis.

There is insufficient evidence to conclude that 5% of children under 18 years of age have asthma.

Part b

P-value = 0.0008

(by using z-table or excel)

Level of significance = = 0.06

P-value < = 0.06

So, we reject the null hypothesis

There is insufficient evidence to conclude that 5% of children under 18 years of age have asthma.

Question 4

Part a

The mean and standard deviation by using calculator are given as below:

Mean = 3991.889

Standard deviation = 224.4273

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.