Question

1. An investigator wants to determine if a new program aimed at reducing the overall number of pregnancies in a highly-populated area is effective (reducing average births below 6 per household). In order to carry out the investigation, the investigator surveys 20 participants in the program and finds the average number of births is 4.7 with a sample standard deviation of 0.9 births. a. Carry out a test of hypothesis to test the investigator’s claim by answering the following with alpha=.01. Research hypothesis: Ha: H0: Test statistic: P-value Conclusion in terms of the problem: Possible error in terms of the problem b. Construct a 95% confidence interval for the average number of births per household in this study. Provide a correct interpretation of your interval. c. How many women should be sampled to estimate the actual average number of births within 1 person and a 96% confidence?

Answer 1

Part a

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H₀: the average number of pregnancies in a highly-populated area is 6.

Alternative hypothesis: H_a: the average number of pregnancies in a highly-populated area is less than 6.

H₀: µ = 6 versus H_a: µ < 6

This is a lower tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 6

Xbar = 4.7

S = 0.9

n = 20

df = n – 1 = 19

α = 0.01

Critical value = -2.5395

(by using t-table or excel)

t = (4.7 – 6)/[0.9/sqrt(20)]

t = -6.4598

P-value = 0.0000

(by using t-table)

P-value < α = 0.01

So, we reject the null hypothesis

There is sufficient evidence to conclude that the average number of pregnancies in a highly-populated area is less than 6.

Part b

Confidence interval for Population mean is given as below:

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

From given data, we have

Xbar = 4.7

S = 0.9

n = 20

df = n – 1 = 19

Confidence level = 95%

Critical t value = 2.0930

(by using t-table)

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

Confidence interval = 4.7 ± 2.0930*0.9/sqrt(20)

Confidence interval = 4.7 ± 0.4212

Lower limit = 4.7 - 0.4212 =4.2788

Upper limit = 4.7 + 0.4212 =5.1212

Confidence interval = (4.2788, 5.1212)

We are 95% confident that the average number of births per household will lies between 4.28 and 5.12.

Part c

We are given

Confidence level = 96%

Critical Z value = 2.0537

(by using z-table)

Margin of error = E = 1

Estimate for σ = 0.9

Sample size formula is given as below:

n = (Z*σ/E)^2

n = (2.0537*0.9/1)^2

n = 3.416324

n = 4