Question

Assume that a researcher wants to collect data to make inference about the hourly wage in New Paltz area. The hourly wage in New Paltz area also follows a normal distribution and the standard deviation is also $5.

• What will be the required sample size, if the researcher wants to achieve a desired margin of error at $1.00 at the significant level of 95%.

• What will be the required sample size, if the researcher wants to achieve a desired margin of error at $1.00 at the significant level of 98%.

• If the researcher collects the following data of the hourly wage,

17.3 18.6 25.7 19.8 28.5 12.6 18.3 16.2

what’s the 95% confidence interval for the population mean estimation?

Is the hourly wage in New Paltz area significantly different from the state average level, at 95% confidence level?

Based on the previous data, what is the value of the test statistic if the researcher wants to test whether the New Paltz average hourly wage is greater than $17? What is the P-value of this test? What is your conclusion regarding the test at 95% significant level?

Answer 1

1)Given , Margin of error = 1.00

for 95% confidence , zc = 1.96

Required sample size =96

2) For 98% confidence , zc = 2.33

Required sample size =138

3)95% confidence interval for population mean is

where

degrees of freedom = n-1= 8-1=7

For 95% confidence with df = 7 , critical value of t is

tc =2.36 ( from t table)

Thus, 95% confidence interval for mean is

=(15.34 , 23.91)

4)  The null and alternative hypothesis

17

> 17

Test statistic

=1.44

df = n-1=8-1=7

P value = 0.0965

Since P value > 0.05

We fail to reject the null hypothesis at 95% confidence level ( 5% significance level)

Note : P value from excel "=T.DIST.RT(1.44,7)"

Calculation:

	x	(x-xbar)^2
	17.3	5.405625
	18.6	1.050625
	25.7	36.905625
	19.8	0.030625
	28.5	78.765625
	12.6	49.350625
	18.3	1.755625
	16.2	11.730625
sum/ss	157	184.995
xbar=sum/8	19.625
s^2=ss/7		26.4278571
s		5.14080316