Question

1. The Wall Street Journal reported that bachelor’s degree recipients with majors in business average starting salaries of $53,900 in 2012 (The Wall Street Journal, March 17, 2014). The results for a sample of 100 business majors receiving a bachelor’s degree in 2013 showed a mean starting salary of $55,144 with a sample standard deviation of $5,200. Conduct a hypothesis test to determine whether the mean starting salary for business majors in 2013 is greater than the mean starting salary in 2012. Use a = .01 as the level of significance

Answer 1

Answer:

We conclude that the mean starting salary for business majors in 2013 is greater than the mean starting salary in 2012.

Step-by-step explanation:

We are given that the Wall Street Journal reported that bachelor’s degree recipients with majors in business average starting salaries of $53,900 in 2012.

A sample of 100 business majors receiving a bachelor’s degree in 2013 showed a mean starting salary of $55,144 with a sample standard deviation of $5,200.

Let $\mu$ = mean starting salary for business majors in 2013.

So, Null Hypothesis, $H_0$ : $\mu$ $\leq$ $53,900     {means that the mean starting salary for business majors in 2013 is smaller than or equal to the mean starting salary in 2012}

Alternate Hypothesis, $H_A$ : $\mu$ > $53,900     {means that the mean starting salary for business majors in 2013 is greater than the mean starting salary in 2012}

The test statistics that would be used here One-sample t-test statistics because we don't know about population standard deviation;

                               T.S. =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample mean starting salary = $55,144

            s = sample standard deviation = $5,200

            n = sample of business majors = 100

So, the test statistics  =  $\frac{55,144-53,900}{\frac{5,200}{\sqrt{100} } }$  ~ $t_9_9$

                                     =  2.392

The value of t-test statistic is 2.392.

Now, at 0.01 significance level the t table gives a critical value of 2.369 at 99 degree of freedom for right-tailed test.

Since our test statistic is more than the critical value of t as 2.392 > 2.369, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean starting salary for business majors in 2013 is greater than the mean starting salary in 2012.

Answer 2

Answer:

The calculated value t = 4.976 > 2.6264 at 0.01 level of significance

Null hypothesis is rejected

Alternative hypothesis is accepted

The  mean starting salary for business majors in 2013 is greater than the mean starting salary in 2012

Step-by-step explanation:

Given Mean of the population μ = $53,900

Given sample size 'n' = 100

Mean of the sample size x⁻ = 55,144

Sample standard deviation 'S' = 5200

Null hypothesis:H₀: There is no difference between the means

Alternative Hypothesis :H₁: The  mean starting salary for business majors in 2013 is greater than the mean starting salary in 2012

Test statistic

$t = \frac{x^{-} -mean}{\frac{S}{\sqrt{n} } }$

$t = \frac{55144-53900}{\frac{5200}{\sqrt{100} } }$

t = 4.976

Degrees of freedom

ν = n-1 = 100-1 =99

t₀.₀₁ = 2.6264

The calculated value t = 4.976 > 2.6264 at 0.01 level of significance

Null hypothesis is rejected

Alternative hypothesis is accepted

Final answer:-

The  mean starting salary for business majors in 2013 is greater than the mean starting salary in 2012