Respuesta :
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
[tex]t = \frac{x^{-} -mean}{\frac{S}{\sqrt{n} } }[/tex]
[tex]t = \frac{55144-53900}{\frac{5200}{\sqrt{100} } }[/tex]
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
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 [tex]\mu[/tex] = mean starting salary for business majors in 2013.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\leq[/tex] $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, [tex]H_A[/tex] : [tex]\mu[/tex] > $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. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean starting salary = $55,144
s = sample standard deviation = $5,200
n = sample of business majors = 100
So, the test statistics = [tex]\frac{55,144-53,900}{\frac{5,200}{\sqrt{100} } }[/tex] ~ [tex]t_9_9[/tex]
= 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.
