Question

the data below represents yearly car insurance premiums in the us for 9 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 $1,576 using a level of significance of 10%.
1,643 1,679 1,497
1,559 1,673 1,487
1,624 1,621 1,492
a. what type of test will be used in this problem? select an answer
b. identify the null and alternative hypotheses?
$h_0$: select an answer?
$h_a$: select an answer?
c. is the original claim located in the null or alternative hypothesis? select an answer
d. calculate your test statistic. write the result below, and be sure to round your final answer to two decimal places.

Explanation:

Step1: Determine test type

Since the population standard - deviation is unknown and we have a sample of data to test a claim about the population mean, a one - sample t - test will be used.

Step2: State hypotheses

The null hypothesis \(H_0\) is the statement of no effect or the claim we assume to be true initially. The alternative hypothesis \(H_a\) is what we are trying to find evidence for. The claim is that the average yearly car insurance premium is \(\mu = 1576\). So, \(H_0:\mu=1576\) and \(H_a:\mu
eq1576\) (a two - tailed test).

Step3: Identify claim location

The original claim \(\mu = 1576\) is located in the null hypothesis \(H_0\).

Step4: Calculate sample mean \(\bar{x}\)

The data values are \(x = \{1643,1679,1497,1559,1673,1487,1624,1621,1492\}\).
\(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\), where \(n = 9\)
\(\sum_{i=1}^{9}x_i=1643 + 1679+1497+1559+1673+1487+1624+1621+1492=14275\)
\(\bar{x}=\frac{14275}{9}\approx1586.11\)

Step5: Calculate sample standard - deviation \(s\)

\[s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\]
First, calculate \((x_i-\bar{x})^2\) for each \(i\):
\((1643 - 1586.11)^2=(56.89)^2 = 3237.47\)
\((1679 - 1586.11)^2=(92.89)^2 = 8627.55\)
\((1497 - 1586.11)^2=(-89.11)^2 = 7941.59\)
\((1559 - 1586.11)^2=(-27.11)^2 = 734.95\)
\((1673 - 1586.11)^2=(86.89)^2 = 7550.87\)
\((1487 - 1586.11)^2=(-99.11)^2 = 9822.79\)
\((1624 - 1586.11)^2=(37.89)^2 = 1435.65\)
\((1621 - 1586.11)^2=(34.89)^2 = 1217.31\)
\((1492 - 1586.11)^2=(-94.11)^2 = 8856.69\)
\(\sum_{i = 1}^{9}(x_i-\bar{x})^2=3237.47+8627.55+7941.59+734.95+7550.87+9822.79+1435.65+1217.31+8856.69 = 49421.97\)
\(s=\sqrt{\frac{49421.97}{9 - 1}}=\sqrt{\frac{49421.97}{8}}\approx78.45\)

Step6: Calculate t - statistic

The formula for the one - sample t - statistic is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\)
Here, \(\bar{x}\approx1586.11\), \(\mu = 1576\), \(s\approx78.45\), and \(n = 9\)
\(t=\frac{1586.11 - 1576}{78.45/\sqrt{9}}=\frac{10.11}{78.45/3}=\frac{10.11}{26.15}\approx0.39\)

Answer:

a. One - sample t - test
b. \(H_0:\mu = 1576\), \(H_a:\mu
eq1576\)
c. Null hypothesis
d. \(0.39\)