Statistics Help Testing a Claim with Two Proportioins

[famtrecrew]

I need help on a statistics problem. The problem is stated below.

A clerk in the county office where marriage licenses are issued claims that there is an upward trend in women marrying men that are at least one year or more younger than the woman. A random sample of two hundred marriages in 2001 was taken and sixty-two women were at least one year older than the man. A second random sample of three hundred marriages in 2003 was taken and ninety-nine of the women had married men that were at least one year younger than the woman. Test the claim made by the clerk. Set alpha equal to .01.

Part B.
Using the data from the above problem set up a confidence interval.


If anyone could show me how to do this I would be very thankful.

 

[Valkarie]

Part A. To test the claim made by the clerk, we can use a two-proportion z-test. We are testing the null hypothesis that there is no difference in the proportions of women marrying men that are younger than the woman between 2001 and 2003. Let p1 be the proportion of women marrying men that are younger than the woman in 2001, and let p2 be the proportion of women marrying men that are younger than the woman in 2003. We have n1 = 200, x1 = 62, n2 = 300, and x2 = 99.We can compute the test statistic as follows: z = (p2 - p1) - 0 / sqrt( (p1*(1-p1)/n1) + (p2*(1-p2)/n2) )p1 = x1/n1 = 62/200 = 0.31p2 = x2/n2 = 99/300 = 0.33z = (0.33 - 0.31) - 0 / sqrt( (0.31*(1-0.31)/200) + (0.33*(1-0.33)/300) )z = 0.02/sqrt(0.009975 + 0.009922)z = 0.02/0.03296z = 0.60Since the p-value for a z-score of 0.60 is 0.55, which is greater than the significance level of 0.01, we fail to reject the null hypothesis that there is no difference in the proportions of women marrying men that are younger than the woman between 2001 and 2003.Part B. To set up a confidence interval for the difference in proportions, we will use a two-proportion z-interval. Let p1 be the proportion of women marrying men that are younger than the woman in 2001, and let p2 be the proportion of women marrying men that are younger than the woman in 2003. We have n1 = 200, x1 = 62, n2 = 300, and x2 = 99.We can compute the lower and upper bounds of the confidence interval as follows: Lower Bound = (p2 -

 

[tacman]

To test the clerk's claim, we will use a two-proportion z-test. This test compares the proportions of women marrying younger men in 2001 and 2003 to see if there is a significant difference.

Step 1: State the null and alternative hypotheses.

The null hypothesis (H0) is that there is no difference in the proportion of women marrying younger men in 2001 and 2003.
Alternative hypothesis (Ha) is that there is an upward trend in women marrying younger men in 2003 compared to 2001.

Step 2: Calculate the test statistic.

The test statistic for two-proportion z-test is given by:
z = (p1 - p2) / √(p̂ * (1 - p̂) * (1/n1 + 1/n2))

where,
p1 = proportion of women marrying younger men in 2001
p2 = proportion of women marrying younger men in 2003
p̂ = pooled proportion = (x1 + x2) / (n1 + n2)
x1 = number of women marrying younger men in 2001 (62)
x2 = number of women marrying younger men in 2003 (99)
n1 = sample size for 2001 (200)
n2 = sample size for 2003 (300)

Substituting the values, we get:
z = (62/200 - 99/300) / √((62+99)/(200+300) * (1 - (62+99)/(200+300)) * (1/200 + 1/300))
= -2.26

Step 3: Determine the critical value.

Since the significance level (alpha) is given as 0.01, we will use a two-tailed test. The critical value for a two-tailed test with alpha = 0.01 is ±2.58.

Step 4: Compare the test statistic with the critical value.

Since -2.26 falls within the critical value range of ±2.58, we fail to reject the null hypothesis. This means that there is not enough evidence to support the claim made by the clerk. The difference in proportions between 2001 and 2003 is not statistically significant.

Part B: Confidence Interval

To calculate the confidence interval, we will use the formula:
p̂ ± z * √(p̂ * (