A research organization claims that the monthly wages of industrial workers in district X

exceeds that of those in district Y by more than Rs 150. Two different samples drawn
independently from the two district yielded the following results:
District X: 1 x = 648, 2
1 s
= 120, and 1 n = 100
District Y: 2 x = 495, 2
2 s = 140, and 2 n = 90
Verify at 0.05 level of significance whether the sample results support the claim of the
organization.

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.

To verify whether the sample results support the claim of the organization, we need to perform a hypothesis test. The null hypothesis (H0) assumes that there is no significant difference in the monthly wages between the two districts. The alternative hypothesis (H1) assumes that the monthly wages in district X exceed district Y by more than Rs 150.

Let's follow these steps to conduct the hypothesis test at a significance level of 0.05:

Step 1: State the hypotheses:
H0: μX - μY ≤ 150 (There is no significant difference in the monthly wages between the two districts)
H1: μX - μY > 150 (The monthly wages in district X exceed district Y by more than Rs 150)

Step 2: Determine the test statistic:
We will use the two-sample t-test to compare the means of the two independent samples.

The formula for the test statistic is given by:
t = (x1 - x2 - d) / sqrt(s1^2/n1 + s2^2/n2)

where:
x1 and x2 are the means of the two samples (648 and 495, respectively)
s1 and s2 are the standard deviations of the two samples (120 and 140, respectively)
n1 and n2 are the sample sizes of the two samples (100 and 90, respectively)
d is the hypothesized difference in means (150)

Step 3: Define the decision rule:
We want to test the alternative hypothesis that the difference in means is greater than 150. For a significance level of 0.05, we will reject the null hypothesis if the test statistic is greater than the critical value from the t-distribution with (n1 + n2 - 2) degrees of freedom.

Step 4: Calculate the test statistic:
Using the formula from Step 2, we can calculate the test statistic:
t = (648 - 495 - 150) / sqrt((120^2 / 100) + (140^2 / 90))
t = 3.480

Step 5: Calculate the critical value:
The critical value can be found using a t-table or a statistical software. For a one-tailed test with a significance level of 0.05 and (n1 + n2 - 2) degrees of freedom, the critical value is approximately 1.666.

Step 6: Make a decision:
Since the test statistic (3.480) is greater than the critical value (1.666), we reject the null hypothesis. This means that there is sufficient evidence to support the claim that the monthly wages in district X exceed district Y by more than Rs 150.

Therefore, the sample results support the claim of the organization at a 0.05 level of significance.