TEST #1 Perform the following two-tailed hypothesis test, using a .05 significance level: Intrinsic by Gender State the null and an alternate statement for the test Use Microsoft Excel (Data Analysis...

TEST #1
Perform the following two-tailed hypothesis test, using a .05 significance level:
 Intrinsic by Gender
 State the null and an alternate statement for the test
 Use Microsoft Excel (Data Analysis Tools) to process your data and run the appropriat
e test. Copy and paste the results of the output to your report in Microsoft Word.
 Identify the significance level, the test statistic, and the critical value.
 State whether you are rejecting or failing to reject the null hypothesis statement.
 Explain how the results could be used by the manager of the company.
Ans: Here we may be interested to know if “Intrinsic” for different genders are same or not.
For that we can test if the population mean of intrinsic for different gender is same or not.
Suppose that µ1 and µ2 are the population intrinsic means for gender 1 and gender 2 respectively.
So the hypothesis of interest is,
H0: µ1=µ2 against H1: µ1≠µ2.
Now we don’t know the value of µ1 and µ2. But since the sample is the best estimator of the pop
ulation so we can use the sample means in this case in case of population mean.
The sample size for both the sample is same, i.e. n=32 (sine n>30 so we can consider this test as
a large sample test). Let ̅ and ̅ are the sample means and
and
are the sample variances f
or the two genders. Now since the samples are independent so the test statistic is,
T=
̅ ̅
√
which follows a t distribution with (2n-2)=62 degrees of freedom under the null hy
pothesis.
And we will reject the null hypothesis if the observed value of the test statistic is greater than the
tabulated critical value, i.e. we reject the null hypothesis at 100α% significance level if,
Observed |T|>tabulated t62,α/2
Obtained table in excel is,
t-Test: Two-Sample Assuming Unequal Variances

Variable
1
Variable
2
Mean 4.9625 5.084375
Variance 0.528226 0.581361
Observations 32 32
Hypothesized Mean
Difference 0
df 62
t Stat -0.6545
P(T<=t) one-tail 0.257606
t Critical one-tail 1.669804
P(T<=t) two-tail 0.515211
t Critical two-tail 1.998972
So in this given question the significance level is 5%, the test statistic is T=-0.6545, the critical valu
e is 1.998972.
Since the absolute value of the test statistic is smaller than the tabulated value so we are failing to reject t
he null hypothesis.
From the obtained result manager of the company can decide that at the 5% level of significance there is
no...