We want to focus on the time for navigation entry. For 24 younger adults (21-36 years), the average time for navigation entry was 31.4 seconds versus 40 seconds for the 24 participants in the older group (55-75 years). Standard deviations are not given, so we estimate the younger group at 0.99 seconds and the older group at 1.04 seconds. Test, using a 5% significance level, to see if there is evidence that the younger group have a lower mean time in seconds than the older group. 1. Write the correct null and alternative hypotheses.

We are given that for 24 younger adults (21-36 years), the average time for navigation entry was 31.4 seconds versus 40 seconds for the 24 participants in the older group (55-75 years).

Standard deviations are not given, so we estimate the younger group at 0.99 seconds and the older group at 1.04 seconds.

We have to conduct a hypothesis test to see whether the younger group have a lower mean time in seconds than the older group.

Let [tex]\mu_1[/tex] = population mean time for navigation entry for younger groups

[tex]\mu_2[/tex] = population mean time for navigation entry for older groups

SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu_1 \geq \mu_2[/tex] or [tex]\mu_1-\mu_2\geq0[/tex] {means that the younger group have a mean time in seconds higher than or equal to the older group}

Alternate Hypothesis, [tex]H_a[/tex] : [tex]\mu_1< \mu_2[/tex] or [tex]\mu_1-\mu_2 < 0[/tex] {means that the younger group have a lower mean time in seconds than the older group}

The test statistics that will be used here is Two-sample t test statistics as we don't know about the population standard deviations;

T.S. = [tex]\frac{(\bar X_1 - \bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1 }{n_1} +\frac{1 }{n_2}} }[/tex] ~ [tex]t__n__1+n_2-_2[/tex]

where, [tex]\bar X_1[/tex] = sample average time for navigation entry for younger adults (21-36 years) = 31.4 seconds

[tex]\bar X_2[/tex] = sample average time for navigation entry for older adults (55-75 years) = 40 seconds

[tex]s_1[/tex] = standard deviation for 24 younger adults = 0.99 seconds

[tex]s_2[/tex] = standard deviation for 24 participants in the older group = 1.04 seconds

[tex]n_1[/tex] = sample size of younger adults = 24

[tex]n_2[/tex] = sample size of older adults = 24

Here, [tex]s_p = \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }[/tex] = [tex]\sqrt{\frac{(24-1)\times 0.99^{2}+(24-1)\times 1.04^{2} }{24+24-2} }[/tex] = 1.015

So, test statistics = [tex]\frac{(31.4-40)-(0)}{1.015 \times \sqrt{\frac{1 }{24} +\frac{1 }{24}} }[/tex] ~ [tex]t_4_6[/tex]

= -29.35

So, at 0.05 level of significance, the t table gives critical value of -1.6792 at 46 degree of freedom for one-tailed test. Since our test statistics is way less than the critical value of t so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the younger group have a lower mean time in seconds than the older group.