Mathematics, 07.04.2020 19:30 heathersloan274

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.

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Answer from: dranio

We conclude that the younger group have a lower mean time in seconds than the older group.

Step-by-step explanation:

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 $\mu_1$ = population mean time for navigation entry for younger groups

$\mu_2$ = population mean time for navigation entry for older groups

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

Alternate Hypothesis, H_a : $\mu_1< \mu_2$ or $\mu_1-\mu_2 < 0$ {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. = $\frac{(\bar X_1 - \bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1 }{n_1} +\frac{1 }{n_2}} }$ ~ t__n__1+n_2-_2

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

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

s_1 = standard deviation for 24 younger adults = 0.99 seconds

s_2 = standard deviation for 24 participants in the older group = 1.04 seconds

n_1 = sample size of younger adults = 24

n_2 = sample size of older adults = 24

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

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

= -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.

Answer from: Quest

gregory's graph is incorrect because it shows 2 inches of growth for each week when it should show 0.75 inches of growth each week.

Answer from: Quest

the correct answer is the second option