in juneau, alaska, the 30 - year annual snowfall average is 86.7 inches with a standard deviation of 40.4…

in juneau, alaska, the 30 - year annual snowfall average is 86.7 inches with a standard deviation of 40.4 inches. the last four years saw an average annual snowfall of 115.7 inches, 62.9 inches, 168.5 inches, and 135.7 inches. hia performs a hypothesis test on this data to determine if the next 30 - year norm will have a different average if the trend from the last four years continues. she uses a significance level of 5%. which of the following is a conclusion that she may make?\nthe z - statistic is 1.44, so the null hypothesis cannot be rejected.\nthe z - statistic is 1.68, so the null hypothesis cannot be rejected.\nthe z - statistic is 1.85, so the null hypothesis should be rejected.\nthe z - statistic is 4.6, so the null hypothesis should be rejected.

in juneau, alaska, the 30 - year annual snowfall average is 86.7 inches with a standard deviation of 40.4 inches. the last four years saw an average annual snowfall of 115.7 inches, 62.9 inches, 168.5 inches, and 135.7 inches. hia performs a hypothesis test on this data to determine if the next 30 - year norm will have a different average if the trend from the last four years continues. she uses a significance level of 5%. which of the following is a conclusion that she may make?\nthe z - statistic is 1.44, so the null hypothesis cannot be rejected.\nthe z - statistic is 1.68, so the null hypothesis cannot be rejected.\nthe z - statistic is 1.85, so the null hypothesis should be rejected.\nthe z - statistic is 4.6, so the null hypothesis should be rejected.

Answer

Explanation:

Step1: Calcular la media muestral

La media muestral $\bar{x}$ de los últimos cuatro años es $\bar{x}=\frac{115.7 + 62.9+168.5 + 135.7}{4}=\frac{482.8}{4}=120.7$

Step2: Calcular la z - estadística

La fórmula para la z - estadística en un test de hipótesis para una media poblacional $\mu$ (con desviación estándar poblacional $\sigma$ y tamaño de muestra $n$) es $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$. Aquí, $\mu = 86.7$, $\sigma=40.4$, $n = 4$ y $\bar{x}=120.7$. $z=\frac{120.7 - 86.7}{\frac{40.4}{\sqrt{4}}}=\frac{34}{\frac{40.4}{2}}=\frac{34}{20.2}\approx1.68$

Step3: Decidir sobre la hipótesis nula

Para un nivel de significancia de $\alpha = 0.05$ en un test de dos colas, los valores críticos son $z_{\alpha/2}=\pm 1.96$. Como $- 1.96<1.68<1.96$, no se rechaza la hipótesis nula.

Answer:

The z - statistic is 1.68, so the null hypothesis cannot be rejected.