car insurance companies want to keep track of the average cost per claim. the current data in use for auto…

car insurance companies want to keep track of the average cost per claim. the current data in use for auto insurance r us is an average of $2,200 for each claim with a standard deviation of $500. with this average, the company can stay competitive with rates but not lose money. however, the statistician for the company believes that the cost of the average claim has increased. he pulled 40 recent claims and found the average to be $2,350. which most restrictive level of significance would suggest that the company should raise rates?\n\n| upper - tail values | | | |\n| ---- | ---- | ---- | ---- |\n| a | 5% | 2.5% | 1% |\n| critical z - values | 1.65 | 1.96 | 2.58 |\n\n1% \n2.5% \n5% \n10%

car insurance companies want to keep track of the average cost per claim. the current data in use for auto insurance r us is an average of $2,200 for each claim with a standard deviation of $500. with this average, the company can stay competitive with rates but not lose money. however, the statistician for the company believes that the cost of the average claim has increased. he pulled 40 recent claims and found the average to be $2,350. which most restrictive level of significance would suggest that the company should raise rates?\n\n| upper - tail values | | | |\n| ---- | ---- | ---- | ---- |\n| a | 5% | 2.5% | 1% |\n| critical z - values | 1.65 | 1.96 | 2.58 |\n\n1% \n2.5% \n5% \n10%

Answer

Explanation:

Step1: Calcular la estadística de prueba z

Usamos la fórmula $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, donde $\bar{x} = 2350$, $\mu = 2200$, $\sigma = 500$ y $n = 40$. $z=\frac{2350 - 2200}{\frac{500}{\sqrt{40}}}=\frac{150}{\frac{500}{6.3246}}\approx\frac{150}{79.0569}\approx1.897$

Step2: Comparar con valores críticos

Tenemos que determinar el nivel de significancia para el cual nuestra $z -$estadística de prueba supera el valor crítico. Vemos que $1.65<1.897<1.96<2.58$.

Answer:

2.5%