the daily temperatures in fall and winter months in virginia have a mean of 62°f. a meteorologist in…

the daily temperatures in fall and winter months in virginia have a mean of 62°f. a meteorologist in southwest virginia believes the mean temperature is colder in this area. the meteorologist takes a random sample of 30 daily temperatures from the fall and winter months over the last five years in southwest virginia. the mean temperature for the sample is 59°f with a standard deviation of 6.21°f. do the data provide convincing evidence at the α = 0.05 level that the mean temperature in fall and winter months in southwest virginia is less than 62°f? what is the test statistic for this significance test? o z = 2.65 o t = 2.65 o z = -2.65 o t = -2.65

the daily temperatures in fall and winter months in virginia have a mean of 62°f. a meteorologist in southwest virginia believes the mean temperature is colder in this area. the meteorologist takes a random sample of 30 daily temperatures from the fall and winter months over the last five years in southwest virginia. the mean temperature for the sample is 59°f with a standard deviation of 6.21°f. do the data provide convincing evidence at the α = 0.05 level that the mean temperature in fall and winter months in southwest virginia is less than 62°f? what is the test statistic for this significance test? o z = 2.65 o t = 2.65 o z = -2.65 o t = -2.65

Answer

Explanation:

Step1: Identify the test - statistic formula

Since the population standard deviation is unknown and we have a sample standard deviation, we use the t - test statistic formula $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $s$ is the sample standard deviation, and $n$ is the sample size.

Step2: Define the values

We are given that $\bar{x} = 59$, $\mu=62$, $s = 6.21$, and $n = 30$.

Step3: Calculate the t - statistic

[ \begin{align*} t&=\frac{\bar{x}-\mu}{s/\sqrt{n}}\ &=\frac{59 - 62}{6.21/\sqrt{30}}\ &=\frac{- 3}{6.21/\sqrt{30}}\ &=\frac{-3}{6.21/5.477}\ &=\frac{-3}{1.134}\ &\approx - 2.65 \end{align*} ]

Answer:

$t=-2.65$