Questions: The data below represents yearly car insurance premiums in the US for 12 randomly selected drivers. Assuming the data is normally distributed, test the claim that the average yearly car insurance premium for a person in the US is 1,566 using a level of significance of 10%. 1,570 1,500 1,385 1,547 1,482 1,616 1,475 1,574 1,615 1,516 1,491 1,652 a. What type of test will be used in this problem? Select an answer b. What evidence justifies the use of this test? Check all that apply: np>5 and nq>5 The original population is approximately normal The sample standard deviation is not known The population standard deviation is known There are two different samples being compared The sample size is larger than 30 The population standard deviation is not known c. Enter the null hypothesis for this test. H0: ? v d. Enter the alternative hypothesis for this test. H1: ?v?v e. Is the original claim located in the null or alternative hypothesis?

The data below represents yearly car insurance premiums in the US for 12 randomly selected drivers. Assuming the data is normally distributed, test the claim that the average yearly car insurance premium for a person in the US is 1,566 using a level of significance of 10%. 1,570 1,500 1,385 1,547 1,482 1,616 1,475 1,574 1,615 1,516 1,491 1,652 a. What type of test will be used in this problem? Select an answer b. What evidence justifies the use of this test? Check all that apply: np>5 and nq>5 The original population is approximately normal The sample standard deviation is not known The population standard deviation is known There are two different samples being compared The sample size is larger than 30 The population standard deviation is not known c. Enter the null hypothesis for this test. H0: ? v d. Enter the alternative hypothesis for this test. H1: ?v?v e. Is the original claim located in the null or alternative hypothesis?
Transcript text: The data below represents yearly car insurance premiums in the US for 12 randomly selected drivers. Assuming the data is normally distributed, test the claim that the average yearly car insurance premium for a person in the US is $\$ 1,566$ using a level of significance of $10 \%$. \begin{tabular}{|l|l|l|} \hline 1,570 & 1,500 & 1,385 \\ \hline 1,547 & 1,482 & 1,616 \\ \hline 1,475 & 1,574 & 1,615 \\ \hline 1,516 & 1,491 & 1,652 \\ \hline \end{tabular} a. What type of test will be used in this problem? Select an answer b. What evidence justifies the use of this test? Check all that apply: $n p>5$ and $n q>5$ The original population is approximately normal The sample standard deviation is not known The population standard deviation is known There are two different samples being compared The sample size is larger than 30 The population standard deviation is not known c. Enter the null hypothesis for this test. $H_{0}$ $\square$ ? $v$ $\square$ $\square$ d. Enter the alternative hypothesis for this test. $H_{1}$ :?v?v $\square$ $\square$ e. Is the original claim located in the null or alternative hypothesis? $\square$
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Solution

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Solution Steps

Step 1: Calculate the Sample Mean

The sample mean \( \bar{x} \) is calculated as follows:

\[ \bar{x} = \frac{\sum_{i=1}^N x_i}{N} = \frac{18423}{12} = 1535.25 \]

Step 2: Calculate the Sample Standard Deviation

The variance \( \sigma^2 \) is calculated using the formula:

\[ \sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} = 5633.66 \]

The standard deviation \( \sigma \) is then:

\[ \sigma = \sqrt{5633.66} = 75.06 \]

Step 3: Calculate the Standard Error

The standard error \( SE \) is calculated as:

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{75.06}{\sqrt{12}} \approx 21.668 \]

Step 4: Calculate the Test Statistic

The test statistic \( t \) is calculated using the formula:

\[ t = \frac{\bar{x} - \mu_0}{SE} = \frac{1535.25 - 1566}{21.668} \approx -1.4191 \]

Step 5: Calculate the P-value

For a two-tailed test, the P-value is calculated as:

\[ P = 2 \times (1 - T(|z|)) \approx 0.1836 \]

Step 6: Identify the Test Type and Hypotheses

The type of test used is a single-sample mean hypothesis test (t-test). The null hypothesis \( H_0 \) and alternative hypothesis \( H_1 \) are defined as follows:

  • Null Hypothesis: \( H_0: \mu = 1566 \)
  • Alternative Hypothesis: \( H_1: \mu \neq 1566 \)

Final Answer

  • a. Test Type: Single-sample mean hypothesis test (t-test)
  • b. Evidence: \(\text{Evidence: } \left[\text{The original population is approximately normal, The population standard deviation is not known}\right]\)
  • c. Null Hypothesis: \( H_0: \text{The average yearly car insurance premium is } \$1,566 \)

Thus, the final answers are: \[ \boxed{\text{a. Test Type: Single-sample mean hypothesis test (t-test)}} \] \[ \boxed{\text{b. Evidence: }\left[\text{The original population is approximately normal, The population standard deviation is not known}\right]} \] \[ \boxed{c. H_0: \text{The average yearly car insurance premium is } \$1,566} \]

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