We hear that listening to Mozart improves students’ performance on tests. Maybe pleasant odors have a similar effect. To test this idea, $21$subjects worked two different but roughly equivalent paper-and-pencil mazes while wearing a mask. The mask was either unscented or carried a floral scent. Each subject used both masks, in a random order. The table below gives the subjects’ times with both masks.

Given in the question that, We hear that listening to Mozart improves students’ performance on tests. Maybe pleasant odors have a similar effect. To test this idea, $21$subjects worked two different but roughly equivalent paper-and-pencil mazes while wearing a mask. The mask was either unscented or carried a floral scent. Each subject used both masks, in a random order. The table below gives the subjects’ times with both masks

The randomly assign the order in which each subject used the two masks.

It is critical to randomly assign the order is made sense of as beneath,

To test for the effectiveness of floral scent on performance, 21 subjects worked two unique yet generally identical paper-and-pencil mazes while wearing the mask.

The mask may be either unscented or scented. Each subject involved both mask in a random order.

Randomization minimizes the contrast between groups by similarly distributing the people. By randomizations people are doled out to relating treatment randomly. Randomization decreases the onlooker inclination, the impact of the request in which the treatments are administered and represents the uncontrolled variables in the experiment.

Thus, irregular task is vital to lessen the baseness.

Given in the question that, We hear that listening to Mozart improves students’ performance on tests. Maybe pleasant odors have a similar effect. To test this idea,$21$subjects worked two different but roughly equivalent paper-and-pencil mazes while wearing a mask. The mask was either unscented or carried a floral scent. Each subject used both masks, in a random order. The table below gives the subjects’ times with both masks

The data provide convincing evidence that the floral scent improved performance. Test whether the floral scent improved performance or not:

The investigator is specifically interested to test whether the floral scent improved the performance or not. Since, $d=$Unscented - Scented, the mean difference (Unscented-Scented) for the rats ${\mu }_{\mathrm{d}}$would be negative if the floral scent is effective

Denote ${\mu }_{\mathrm{d}}$as the population mean difference in the performance with scented and unscented.

The null and alternate hypotheses are stated below:

Null hypothesis ${\mathrm{H}}_0$:

${\mathrm{H}}_0:{\mu }_{\mathrm{d}}=0$

That is, there is no significant difference in the mean performance with scented masks and unscented masks.

${\mathrm{H}}_1:{\mu }_{\mathrm{d}}<0$

That is, the mean performance with scented masks is significantly greater than the mean performance with unscented masks.

In order to test the hypothesis regarding the significant difference between a paired set of $n$observations $\left(x_i,y_i\right)$, a paired t-test is appropriate.

Thus, the sampling distribution used in the given scenario is "Student's $t^{\text{'}\text{'}}$by assuming that $d$has an approximately normal distribution.

Paired t test:

EXCEL software can be used to perform a paired $t$test for the given data.

Software Procedure:

The step-by-step procedure to perform a paired t test using EXCEL software is given below:

- Open an EXCEL file.

- Enter the data of Unscented in column A and name it as Unscented.

- Enter the data of Scented in column B and name it as Scented.

- Select Data > Data Analysis > t-Test: Paired Two Sample for Means > OK.

- In Variable 1 Range enter \localid="1650366688716" $AS1:\backslash$AS22 and in Variable 2 Range enter $\backslash$B \localid="1650366702969" $1:\backslash$B \localid="1650366696932" $22.$

- Click on Labels and enter Alpha aslocalid="1650366709523" $05.$

- Click OK.

The obtained output is shown below

From the above obtained output below values are reported:

The test statistic is $t=0.3494$.

The degrees of freedom is d.f $=20$.

P-value:

From the above obtained output, the P-value for right tailed test is P-value $=0.3652$.

Since, he hypothesis test is Left tailed, the P-value is P-value $=1-0.3652=0.6348$.

Thus, the P-value is P-value $=0.6348$.

Decision rule:

The level of significance is $a=0.05$.

Decision rule based on P-value approach:

If $\mathrm{P}$-value $\le \mathrm{a}$, then reject the null hypothesis ${\mathrm{H}}_0$.

If $P$-value$>a$, then fail to reject the null hypothesis ${\mathrm{H}}_0$.

Conclusion:

Conclusion based on P-value approach:

The$P$-value is $0.6348$and a value is $0.05$.

Here, P-value is greater than the a value.

That is, $0.6348$(=P-value)$>0.05$ (=a).

By the rejection rule, fail to reject the null hypothesis.

Thus, there is no significant difference in the mean performance with scented masks and unscented masks.

Hence, at $0.05$level of significance, there is not enough evidence to conclude that the floral scent improved performance.