Question

Elementary Statistics Name: Study Guide 27 a Class: Due Date: Score: No Work No Polnts Use Pencil Only Be Neat & Organized 1. A local nurse's union has done a study on salary of full-time nurses, The result of this study is summarized in the table below. Females Males 60 ng48. ,-7050 36750 ir 81 Not Given sa Not Given a-275 o2250 Table 1: Monthly Salaries For Nurses (a) (3 points) Construct a 98% confidence interval for the difference between pop- ulation means -2 using data in table 1. (a) (b) (2 points) Compute the margin of error. (b) A local newspaper claims that the mean salary of all full-time female nurses is more than the mean salary of all full-time male nurses. Test this claim at a 0.02 by using the data in table 1. Total Points: 50 Study Guide 27 Page 1 of 4

Answer 1

Solution:-

1)

a) 98% confidence interval for the difference in means is C.I = ( 182.205, 417.795).

2 C.I= (피-12) ± za/2 X n2 ni

C.I = (7050 - 6750) + 2.327 × 50.6211

C.I = 300 + 117.7954

C.I = ( 182.205, 417.795)

b) Margin of error is 117.7954.

2 M.E tza/2 X n2 ni

M.E = + 2.327 × 50.6211

M.E = + 117.7954

c)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_Female< u_Male
Alternative hypothesis: u_Female > u_Male

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.02. Using sample data, we will conduct a two-sample z-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), Z statistic test statistic (z).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 50.6211

z_Critical = + 2.054

Rejection region is z > 2.054

p 0.02 -3 2 -1 0 3 z 2.054

e)

z = [ (x₁ - x₂) - d ] / SE

z = 5.93

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is thesize of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of 5.93

Therefore, the P-value in this analysis is less than 0.0001

Interpret results. Since the z-value lies in the rejection region, hence we have to reject the null hypothesis.

f) From the above test we can conclude that the mean salary of all full time females is more than the mean salary of all full time male nurses.