91Ó°ÊÓ

The significance level is $5\%$

$x=394$

$n=758$

The sample proportion is expressed as

$\hat{p}=\frac{x}{n}$

The sample proportion is calculated as

$\hat{p}=\frac{394}{758}$

$=0.52$

As a result, the sample proportion is $0.52$, indicating that the$52\%$ majority of Arizona families who celebrate Christmas wait until the day after the holiday to open their gifts.

The significance level is $5\%$

$x=394$

$n=758$

Calculate the value of$n{p}_0$

$n{p}_0=\left(758\right)(0.5)$

$=379$

Calculate the value of$n\left(1-p_0\right)$

$n\left(1-p_0\right)=\left(758\right)(1-0.5)$

$=758(0.5)$

$=379$

Both the values are greater than $5$.

So one proportion $z-$test is appropriate to use.

The null hypothesis is

$H_0:p_0=0.5$

The alternate hypothesis is

$H_0:p_0>0.5$

The expression for $z$is

$z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}$

The z value is calculated as

$z=\frac{0.52-0.5}{\sqrt{\frac{0.5(1-0.5)}{758}}}$

$=\frac{0.02}{0.018}$

$=1.09$

And also $\mathrm{Î\pm }=0.05$, the critical value of $z$from the standard table is

$z_{0.05}=1.645$

Because the test is a right-tailed test, a positive value is used.

The test statistic is within acceptable bounds. As a result, the hypothesis $H_0$is not ruled out.

At the $5\%$level, the test results are not statistically significant.

As a result, the data does not support the conclusion that the majority of Arizona Christmas-observant families wait until Christmas Day to unwrap their gifts.