Math  /  Data & Statistics

QuestionIn a ready to serve dinner preparation study of 10 trials of a new meal, the sample mean time for preparation was 18.5 min . and a sample standard deviation of 5.9 min . Prep time can be assumed normally distributed. Test whether the new meal meets the standard for having a preparation time below the 20 min . standard. Use 1%1 \% significance level.
What is the CONCLUSION for the study if HO: mean=20 min.? 1) Reject HO and conclude preparation time is below 20 min . 2) Fail to Reject HO and conclude the preparation time is below 20 min . 3) Reject HO and conclude the preparation time does not differ from 20 min . 4) Fail to Reject HO and conclude the preparation time does not differ from 20 min . 5) Fail to Reject the alternative H 1 and conclude HO is true. 6) No answer is correct.

Studdy Solution

STEP 1

What is this asking? Does the new ready-to-eat meal take less than 20 minutes to prepare, based on our tests? Watch out! Don't mix up "reject" and "fail to reject," and make sure you're looking at the right significance level!

STEP 2

1. Set up the hypothesis test
2. Calculate the test statistic
3. Find the critical value
4. Make a decision

STEP 3

We're testing if the mean prep time is less than 20 minutes.
Our **null hypothesis** (H0H_0) is that the mean prep time *is* 20 minutes: H0:μ=20H_0: \mu = 20.

STEP 4

Our **alternative hypothesis** (H1H_1) is that the mean prep time is *less than* 20 minutes: H1:μ<20H_1: \mu < 20.
This is a *one-tailed test* because we're only looking at one direction (less than).

STEP 5

Our **significance level** (α\alpha) is α=0.01\alpha = 0.01, or 1%1\%.
This means there's a 1%1\% chance we'll reject the null hypothesis even if it's true.

STEP 6

We'll use a *t-test* because we have a small sample size (10 trials) and we don't know the population standard deviation.
The formula for the t-statistic is: t=xˉμsnt = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} where xˉ\bar{x} is the **sample mean**, μ\mu is the **population mean** (from our null hypothesis), ss is the **sample standard deviation**, and nn is the **sample size**.

STEP 7

Let's plug in our values: xˉ=18.5\bar{x} = 18.5, μ=20\mu = 20, s=5.9s = 5.9, and n=10n = 10. t=18.5205.910t = \frac{18.5 - 20}{\frac{5.9}{\sqrt{10}}}

STEP 8

Now we calculate! t=1.55.93.162=1.51.8660.798t = \frac{-1.5}{\frac{5.9}{3.162}} = \frac{-1.5}{1.866} \approx -0.798 So, our **t-statistic** is approximately 0.798-0.798.

STEP 9

Since this is a one-tailed test with a significance level of 0.010.01 and 99 degrees of freedom (n1=101=9n - 1 = 10 - 1 = 9), we need to find the critical t-value that corresponds to these values.
We can look this up in a t-table or use a calculator.

STEP 10

For a one-tailed test with α=0.01\alpha = 0.01 and 99 degrees of freedom, the **critical t-value** is approximately 2.821-2.821.
We're looking at the left tail of the distribution because our alternative hypothesis is μ<20\mu < 20.

STEP 11

Our **t-statistic** (0.798-0.798) is greater than our **critical t-value** (2.821-2.821).

STEP 12

Because our t-statistic is *greater* than the critical t-value, we *fail to reject* the null hypothesis.
In simpler terms, we don't have enough evidence to say the mean prep time is less than 20 minutes.

STEP 13

The correct answer is 4) Fail to Reject HO and conclude the preparation time does not differ from 20 min.

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