Question

Bernard ran an experiment to test optimum power and time settings for microwave popcorn. His goal was to deliver popcorn with fewer than 7​% of the kernels left​ unpopped, on average. He determined that power 9 at 4 minutes was the best combination. To be sure that the method was​ successful, he popped 8 more bags of popcorn​ (selected at​ random) at this setting. All were of high​ quality, with the percentages of unpopped kernels shown below.

5.5​,4.8​,11.5​,2.3​,2.5​,6.8​,4.1​,6.8  

Does this provide evidence that he met his goal of an average of fewer than 7​% unpopped​ kernels? Use 0.05 as the​ P-value cutoff level.

Calculate the test statistic

t= ____

p-value= _____

Answer 1

Given $\alpha=0.05$

Desciptive Measures:

Finding Mean:

$\overline{x}=\frac{5.5+4.8+11.5+2.3+2.5+6.8+4.1+6.8}{8}=5.5375$

Finding Standard Deviation:

(5.5375 – 5.5)2 +...+ (5.5375 – 6.8)2 8 – 1 ង 2.95341

Hypothesis Test:

Given Claim: Mean should be less than 7%

Null Hypothesis states that the population mean is Equal to the value mentioned in the claim.
If the Null Hypothesis is the claim then the Opposite Hypothesis states the Opposite of the Null Hypothesis.

$H_0:\mu=7$

$H_a:\mu<7$

Determine the value of the Test Statistic :

$t=\frac{\overline{x}-\mu_0}{\frac{s}{\sqrt{n}}}$

$t=\frac{5.5375-7}{\frac{2.95341}{\sqrt{8}}}$

$t=-1.401$

We know that df = 8-1 = 7

Now, Looking up a T-Table, in the Appendix that contains the t-value in the row df (One-tail Probability):

0.01 < P < 0.102

P value is more than Significance Level

So, null hypothesis cannot be rejected.

So, we have Sufficient Evidence that he Did NOT met his requirements.

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