Math  /  Data & Statistics

QuestionA sample of 35 students had mean cell phone age of 25 months with standard deviation 5 . Ten professors had a mean cell phone age of 30 months with standard deviation 8 .
Find the 99%99 \% confidence interval for the diffence in population means for the ages of cell phones. \qquad \square )
Is there sufficient evidence to conclude the age of student and professor cell phones is different? Choose one
Test Statistic: T=T= \square

Studdy Solution

STEP 1

What is this asking? We want to find out if there's a real difference between how old students' cell phones are compared to professors', and how confident we are about that difference. Watch out! Don't mix up the standard deviation of the *sample* with the standard error of the *mean*!

STEP 2

1. Calculate the Standard Error
2. Calculate the Degrees of Freedom
3. Find the Critical T-Value
4. Calculate the Confidence Interval
5. Perform the Hypothesis Test

STEP 3

Let's **start** by calculating the standard error of the difference in means.
This tells us how much the difference between the sample means could vary just by random chance.
The formula is: SE=s12n1+s22n2SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} where s1s_1 and s2s_2 are the standard deviations of the two samples, and n1n_1 and n2n_2 are the sample sizes.

STEP 4

Let's **plug in** the values from the problem: s1=5s_1 = \textbf{5} (students' standard deviation), n1=35n_1 = \textbf{35} (number of students), s2=8s_2 = \textbf{8} (professors' standard deviation), and n2=10n_2 = \textbf{10} (number of professors).

STEP 5

So, the standard error is: SE=5235+8210=2535+6410=57+325=25+22435=249357.1142.667SE = \sqrt{\frac{\textbf{5}^2}{\textbf{35}} + \frac{\textbf{8}^2}{\textbf{10}}} = \sqrt{\frac{25}{35} + \frac{64}{10}} = \sqrt{\frac{5}{7} + \frac{32}{5}} = \sqrt{\frac{25 + 224}{35}} = \sqrt{\frac{249}{35}} \approx \sqrt{7.114} \approx \textbf{2.667}

STEP 6

Next, we need the degrees of freedom.
This helps us find the right t-value for our confidence interval.
A slightly simplified formula for degrees of freedom is: df=(s12n1+s22n2)2(s12/n1)2n11+(s22/n2)2n21df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(s_1^2/n_1)^2}{n_1 - 1} + \frac{(s_2^2/n_2)^2}{n_2 - 1}}

STEP 7

We already calculated the numerator in the standard error calculation!
It's 7.114\textbf{7.114}.
Now for the denominator: (25/35)234+(64/10)29=(5/7)234+(32/5)290.51034+40.9690.015+4.5514.566\frac{(25/35)^2}{34} + \frac{(64/10)^2}{9} = \frac{(5/7)^2}{34} + \frac{(32/5)^2}{9} \approx \frac{0.510}{34} + \frac{40.96}{9} \approx 0.015 + 4.551 \approx 4.566

STEP 8

So, the degrees of freedom are: df=7.11424.56650.614.56611.0811df = \frac{7.114^2}{4.566} \approx \frac{50.61}{4.566} \approx \textbf{11.08} \approx 11

STEP 9

For a 99% confidence interval with 11 degrees of freedom, we look up the critical t-value.
Using a t-table, we find that t0.005,113.106t_{0.005, 11} \approx \textbf{3.106}.

STEP 10

The confidence interval formula is: (xˉ1xˉ2)±tSE(\bar{x}_1 - \bar{x}_2) \pm t \cdot SE where xˉ1\bar{x}_1 and xˉ2\bar{x}_2 are the sample means, tt is the critical t-value, and SESE is the standard error.

STEP 11

Plugging in our values (xˉ1=25\bar{x}_1 = \textbf{25}, xˉ2=30\bar{x}_2 = \textbf{30}, t=3.106t = \textbf{3.106}, SE=2.667SE = \textbf{2.667}): (2530)±3.1062.667(25 - 30) \pm 3.106 \cdot 2.667 5±8.29-5 \pm 8.29So, the 99% confidence interval is approximately (13.29,3.29)(-13.29, 3.29).

STEP 12

Our test statistic is: T=(xˉ1xˉ2)SE=25302.667=52.667-1.87T = \frac{(\bar{x}_1 - \bar{x}_2)}{SE} = \frac{25 - 30}{2.667} = \frac{-5}{2.667} \approx \textbf{-1.87}

STEP 13

Since 1.87-1.87 is within our confidence interval (13.29,3.29)(-13.29, 3.29), which *includes zero*, we *fail to reject* the null hypothesis.
There is *not* sufficient evidence to conclude that the average cell phone ages are different.

STEP 14

99% Confidence Interval: (-13.29, 3.29) Test Statistic: T = -1.87 Not sufficient evidence to conclude a difference in cell phone ages.

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