The students at Littlewood Regional High School cut an average of 3.3 classes per week. A random sample of 117 seniors averages 3.8 cuts per week, with a standard deviation of 0.53. Are seniors significantly different from the student body as a whole?

Answer is 10.2 rounded

Formula is Z over X-bar minus Mue (sp?) OVER Sigma divided by square root of N. H-0 = .57 H-1 not equal to .57

To determine if seniors are significantly different from the student body as a whole, we can perform a hypothesis test.

Let's define our null and alternative hypotheses:

Null Hypothesis (H0): The average number of cuts among seniors is equal to the average number of cuts among the student body as a whole.
Alternative Hypothesis (Ha): The average number of cuts among seniors is significantly different from the average number of cuts among the student body as a whole.

We can use a t-test to test our hypotheses.

First, we need to calculate the test statistic (t-value):

t-value = (sample mean - population mean) / (sample standard deviation / √n)

Where:
- sample mean is 3.8 (average number of cuts among seniors),
- population mean is 3.3 (average number of cuts among the student body as a whole),
- sample standard deviation is 0.53,
- n is the sample size, which is 117.

Let's calculate the t-value:

t-value = (3.8 - 3.3) / (0.53 / √117)

= 0.5 / (0.53 / √117)

Next, we need to find the critical value corresponding to our significance level (usually α = 0.05 for a 95% confidence interval) and degrees of freedom (df), which is the sample size minus one (n-1).

We can then compare the t-value to the critical value to determine if seniors are significantly different from the student body as a whole. If the t-value is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

After performing these calculations, if you provide the degrees of freedom (df), I can help you find the critical value and further evaluate the results.

To determine if the seniors at Littlewood Regional High School are significantly different from the student body as a whole in terms of cutting classes, we can conduct a hypothesis test using the sample mean and standard deviation.

Here are the steps to perform the hypothesis test:

Step 1: Set up the hypotheses:
- Null hypothesis (H0): The average number of cuts per week for seniors is the same as the student body as a whole. μsenior = μstudent
- Alternative hypothesis (Ha): The average number of cuts per week for seniors is significantly different from the student body as a whole. μsenior ≠ μstudent

Step 2: Choose the significance level (α):
The significance level determines the threshold for accepting or rejecting the null hypothesis. Let's assume a significance level of 0.05, which is a common choice.

Step 3: Collect and analyze the data:
In this case, we already have the sample mean and standard deviation for the seniors:
- Sample size (n) = 117
- Sample mean (x̄) = 3.8 cuts per week
- Sample standard deviation (s) = 0.53

Step 4: Calculate the test statistic:
We will use the independent two-sample t-test since we are comparing the means of two independent samples (seniors vs. whole student body). The formula to calculate the test statistic is:

t = (x̄senior - x̄student) / √((s^2senior / nsenior) + (s^2student / nstudent))

Where:
- x̄senior = sample mean of seniors
- x̄student = population mean (3.3 cuts per week for the entire student body)
- s^2senior = sample variance of seniors (variance = standard deviation squared, so s^2senior = 0.53^2)
- nsenior = sample size of seniors (117)
- s^2student = population variance (variance = standard deviation squared, so s^2student = 0.53^2)
- nstudent = total number of students in the whole student body (unknown)

Step 5: Determine the critical value(s):
To find the critical value(s), we need the degrees of freedom (df) for the t-distribution. Since the population standard deviation is unknown, we use the smaller of n1-1 and n2-1. In this case, it is (117-1) = 116.
We can use a t-table or statistical software to find the critical value(s) corresponding to our chosen significance level (α = 0.05) and degrees of freedom (df = 116). Let's assume the critical values are -1.96 and 1.96 (for a two-tailed test).

Step 6: Calculate the p-value:
Using the test statistic we calculated earlier, we can now calculate the p-value. The p-value is the probability of obtaining a test statistic as extreme as the one observed, assuming the null hypothesis is true. We can use the t-distribution to find the p-value.

Step 7: Make a decision:
- If the p-value is less than the significance level (α), we reject the null hypothesis. This indicates that seniors are significantly different from the student body as a whole in terms of cutting classes.
- If the p-value is greater than the significance level (α), we fail to reject the null hypothesis. We do not have enough evidence to conclude that seniors are significantly different from the student body as a whole in terms of cutting classes.

Remember, the p-value represents the likelihood of observing the sample data if the null hypothesis is true.

I can calculate the test statistic and p-value for you if you provide me with the sample size and population mean (x̄student) of the entire student body.

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