a company claims that its packages of beads contain, on average, 50 beads with a standard deviation of 5.4…

a company claims that its packages of beads contain, on average, 50 beads with a standard deviation of 5.4 beads. in a hypothesis test of this claim, $h_0$ is $mu = 50$ and $h_a$ is $mu\neq50$, where $mu$ is the average number of beads per package. based on a sample of 20 packages, celia calculates a mean of 52.5 beads per package. formula for z - statistic: $z=\frac{\bar{x}-mu}{(\frac{sigma}{sqrt{n}})}$ what is the z - statistic for the sample? round the answer to the nearest hundredth. $z = square$ done

a company claims that its packages of beads contain, on average, 50 beads with a standard deviation of 5.4 beads. in a hypothesis test of this claim, $h_0$ is $mu = 50$ and $h_a$ is $mu\neq50$, where $mu$ is the average number of beads per package. based on a sample of 20 packages, celia calculates a mean of 52.5 beads per package. formula for z - statistic: $z=\frac{\bar{x}-mu}{(\frac{sigma}{sqrt{n}})}$ what is the z - statistic for the sample? round the answer to the nearest hundredth. $z = square$ done

Answer

Explanation:

Step1: Identify the values

$\bar{x}=52.5$, $\mu = 50$, $\sigma=5.4$, $n = 20$

Step2: Substitute values into formula

$z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{52.5 - 50}{\frac{5.4}{\sqrt{20}}}$ First, calculate $\sqrt{20}\approx4.472$. Then $\frac{5.4}{4.472}\approx1.207$. And $\frac{52.5 - 50}{1.207}=\frac{2.5}{1.207}\approx2.07$

Answer:

$2.07$