Question38 married couples without children are asked to report the number of times per year they initiate a date night. The men report initiating an average of 8.3 date nights with a standard deviation of 4.1, though it's possible they're overstating to make themselves look good. Is there significant evidence to conclude that married men without children initiate date night 7 times per year at the 0.05 significance level? Note that there's evidence that this distribution is skewed.
What are the hypotheses?
vs
vs
vs
vs
What distribution does the test statistic follow?
t with 39 degrees of freedom
t with 37 degrees of freedom
t with 38 degrees of freedom
z
What is the value of the test statistic? Round to two decimal places. 1.96
What is probability statement for the p-value?
None of the above
Studdy Solution
STEP 1
What is this asking?
Do men without kids plan dates more than 7 times a year, and can we be confident about this, even if the data might be a little wonky?
Watch out!
The men reported the numbers themselves, so they might be exaggerating a bit!
Also, the data isn't perfectly bell-shaped, so we need to use the right kind of test.
STEP 2
1. Set up the Hypothesis Test
2. Calculate the Test Statistic
3. Find the p-value
4. Make a Decision
STEP 3
We're checking if the *true average* number of date nights is greater than 7.
Our **null hypothesis** is that it's not greater than 7, written as .
The **alternative hypothesis**, what we're trying to prove, is .
STEP 4
We're told the **significance level** is 0.05.
This means we're willing to accept a 5% chance of being wrong if we reject the null hypothesis.
STEP 5
Since the data is a little skewed and we don't know the true population standard deviation, we'll use a *t-test*.
The formula for the **t-statistic** is:
Where:
is the **sample mean** (8.3)
is the **hypothesized population mean** (7)
is the **sample standard deviation** (4.1)
is the **sample size** (38)
STEP 6
Let's plug in the numbers! Our **test statistic** is approximately **1.95**.
STEP 7
Because our alternative hypothesis is , we're looking for the probability of getting a t-statistic *greater than* or *equal to* our calculated value, assuming the null hypothesis is true.
Since we're using a *t-distribution*, and our sample size is 38, we have **degrees of freedom**.
The p-value is represented as , where T follows a t-distribution with 37 degrees of freedom.
STEP 8
Using a t-table or calculator with our **test statistic** of approximately **1.95** and **37 degrees of freedom**, we find a p-value of approximately **0.029**.
STEP 9
Our **p-value** (approximately **0.029**) is *less than* our **significance level** of **0.05**.
This means we *reject* the null hypothesis!
STEP 10
The hypotheses are vs .
The test statistic follows a t-distribution with 37 degrees of freedom.
The test statistic is approximately 1.95.
The probability statement for the p-value is .
We reject the null hypothesis and conclude there is significant evidence that married men without children initiate date night more than 7 times per year.
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