L (Valor \( 40 \% \) ) Si se sabe que la dureza Rockwell de pernos de cierto tipo tiene un valor medio de 50 y varianza de 16.44 . 1. Si la distribución es normal, écuál es la probabilidad de que la dureza muestral media para una muestra aleatoria de 9 pernos sea por lo menos 51 ? 2. ¿Cuál es la probabilidad de que la dureza muestral media para una muestra aleatoria de 40 pernos sea no mayor a 52 ?

The first step involved in calculating the Pearson correlation coefficient is to sum all the scores square all the scores calculate the standard deviation for each variable

9 A \( 99 \% \) confidence interval for the mean mass of steel sheets produced in a factory is \( (30.37 \mathrm{~kg}, 32.43 \mathrm{~kg}) \). This interval is based on a random sample of 36 steel sheets. Find a \( 95 \% \) confidence interval for the mean mass of steel sheets, assuming that the masses of steel sheets are normally distributed.

Many investors and financial analysts believe the Dow Jones Industrial Average (DJIA) gives a good barometer of the overall stock market. On January \( 31,2006,9 \) of the 30 stocks making up the DJIA increased in price (The Wall Street Journal, February 1, 2006). On the basis of this fact, a financial analyst claims we can assume that \( 30 \% \) of the stocks traded on the New York Stock Exchange (NYSE) went up the same day. A sample of 76 stocks traded on the NYSE that day showed that 10 went up. You are conducting a study to see if the proportion of stocks that went up is is significantly less than 0.3 . You use a significance level of \( \alpha=\ldots 2 \). What is the test statistic for this sample? Use \( \hat{p} \) rounded to 4 decimal places. (Report answer accurate to 4 decimal places.) test statistic =

Test the claim that the proportion of people who own cats is significantly different than \( 10 \% \) at the 0.2 significance level. The null and alternative hypothesis would be: \( H_{0}: p=0.1 \quad H_{0}: p=0.1 \quad H_{0}: \mu=0.1 \quad H_{0}: \mu=0.1 \quad H_{0}: p=0.1 \quad H_{0}: \mu=0.1 \) \( H_{1}: p \neq 0.1 \quad H_{1}: p<0.1 \quad H_{1}: \mu \neq 0.1 \quad H_{1}: \mu>0.1 \quad H_{1}: p>0.1 H_{1}: \mu<0.1 \) The test is: right-tailed left-tailed two-tailed Based on a sample of 700 people, \( 8 \% \) owned cats The test statistic is: \( -\quad \) This is not a decimal or integer value. (to 4 decimal places) The p-value is:

Se obtuvieron los siguientes resultados para el punto de ebullición, en grados Celsius, de un compuesto de silicio: \( \mathbf{1 6 6} \), \( \mathbf{1 4 1 , 1 3 6 , 1 5 3 , 1 7 0 , 1 6 2 , 1 5 5 , 1 4 6 , 1 8 3 ,} \) \( \mathbf{1 5 7 , 1 4 8 , 1 3 2 , 1 6 0 , 1 7 5 y} \mathbf{1 5 0} \). Utilice la prueba de Kolmogorov- Smirnoff con un nivel de significancia de \( \mathbf{0 . 0 1} \) para probar la hipótesis nula de que los puntos de ebullición provienen de una población normal con una media \( \boldsymbol{\mu = 1 6 0}{ }^{\circ} \mathrm{C} \) y una desviación estándar \( \boldsymbol{\sigma = 1 0 ^ { \circ }} \mathrm{C} \)