Question

After seeing countless commercials claiming one can get cheaper car insurance from an online company, a local insurance agent was concerned that he might lose some customers. To investigate, he randomly selected profiles for 10 of his clients and checked online price quotes for their policies. Then, based on paired data on the 10 clients (the price he offered them and the corresponding online quote) and using Excel’s t-test for paired data, he obtained t-statistic = 0.709 p-value = 0.248 Note that the above is based on the difference between his price and online quotes.

What are the null and alternative hypotheses, respectively?

Answer 1

Null Hypothesis,
`H_0` :
`\mu_d` = 0

Alternate Hypothesis,
`\mu_d\\eq` 0

Explanation:

We are given that a local insurance agent randomly selected profiles for 10 of his clients and checked online price quotes for their policies.

Then, based on paired data on the 10 clients (the price he offered them and the corresponding online quote), he obtained t-statistic = 0.709 p-value = 0.248.

Here , we will use the concept of Paired data test statistics because the prices that the local insurance agent offered and the corresponding online quotes are from the same single data. These information has not been taken from the two independent samples.

So, Null Hypothesis,
`H_0` :
`\mu_d` = 0

Alternate Hypothesis,
`\mu_d\\eq` 0

Here, null hypothesis states that there is no difference between the price he offered them and the corresponding online quote.

On the other hand, alternate hypothesis states that there is difference between the price he offered them and the corresponding online quote.

Hence, this would be the correct null and alternative hypothesis.