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 \( \mu \neq 50 \), 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: \[ z=\frac{\bar{x}-\mu}{\left(\frac{\sigma}{\sqrt{n}}\right)} \] What is the z-statistic for the sample? Round the answer to the nearest hundredth. \[ z=2.07 \] COMPLETE The critical value for \( z^{\star} \) for a hypothesis test of the claim at \( 5 \% \) significance is \( z^{\star}=1.96 \). How should the \( z \)-statistic for the sample be interpreted in terms of the hypothesis test? The null hypothesis should be rejected. The null hypothesis should be accepted. There is not enough evidence to reject the null hypothesis.
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When the z-statistic is calculated as \( z = 2.07 \), it indicates that the observed sample mean of 52.5 beads is 2.07 standard deviations away from the hypothesized population mean of 50 beads. Since this calculated z-statistic exceeds the critical value of \( z^{\star} = 1.96 \) for a two-tailed test at a 5% significance level, it provides strong evidence against the null hypothesis. In practical terms, a z-statistic above the critical threshold means that there is a significant difference between the observed and hypothesized means, suggesting that the company’s claim may not hold. Thus, it leads to the conclusion that the null hypothesis should be rejected, indicating that there is statistically significant evidence to suggest the average number of beads per package is indeed different from the company's claim.
