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 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 = population mean time for navigation entry for younger groups

      = population mean time for navigation entry for older groups

SO, Null Hypothesis, :    or    {means that the younger group have a mean time in seconds higher than or equal to the older group}

Alternate Hypothesis, :  or    {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. =  ~  

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

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

  = standard deviation for 24 younger adults = 0.99 seconds

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

 = sample size of younger adults = 24

  = sample size of older adults = 24

Here, = = 1.015

So, test statistics =    ~

                               = -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.

the correct answer is $7.38

step-by-step explanation: