Question

An insurance provider states that their customers save at least, on average, 300 dollars per year by switching to them, with a standard deviation of 150 dollars. Before we decide to switch to the new company and go through all of the hassle, we want to test the claim. So, we go out and sample 64 individuals who switched to the new insurance company and found them to have saved an average of 255 dollars per year. Do we have enough evidence at the α = 0.05 level to state that the insurance provider is false in their claim? Discussion Prompts Answer the following questions in your initial post: What are the hypotheses based on the words given in the problem? Should we use a Z or T distribution in this case? What is our Z or T statistic? What is the P-value? Based on your p-value and alpha, what conclusion will we make? Based on your results, would you switch to this company? Explain why or why not (Note: this can go beyond the use of statistics, but statistical analysis can help our decisions)

Answer 1

The solution to the question is mathematically given as

1)

H0: M $\geqslant$ 300

H0: M $\geqslant$ 300

2)

Z distribution.

3)

z=-2.4

4)

P=0.0082

5)

"H0" is rejected as a hypothesis with a level of significance of 0.05.

What is the hypothesis ?

Generally, the equation for is mathematically given as

Solutions

we have, $u=300 \begin{aligned}& \sigma=150 \text { dollars } \\N &=64 \text { individuals } \\\bar{x} &=255 \text { dollare. } \\a=& 0.05 .\end{aligned}$

(1):

H0: M $\geqslant$ 300 dollars;  customers save at least 300 dollars per year by switching to them.

H0: M $\geqslant$ 300 dollars:

(This is a left-tailed test)

(2):

when $\sigma$ is known, we use the Z test. we use Z distribution.

(3):

$\begin{gathered}z=\frac{\bar{x}-\mu}{6 / \sqrt{n}}=\frac{255-300}{150 / \sqrt{64}}=\frac{-45}{18.75}=-2.4 \\z=-2.4\end{gathered}$

(4):

$Pvalue =P(z < -2 \cdot 4) =0.0082 \quad\{ wing z tables \} \\\\\text { Pvalue }=0.0082$

(5):

In conclusion, Since p-value = 0·0082 <0.05

This is statistically significant at the 0.05 level.

We thus "Refect H0" using a threshold of significance of 0.05.

"H0" is rejected as a hypothesis with a level of significance of 0.05.

There is not enough data to support the assertion that consumers may save at least $300 annually by switching insurance providers.

Therefore, we would not consider making the transition to this firm.

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