Please see the attached photo. I do not know how to calculate any of this using my TI 84+ CE calculator and the answers I am getting when trying to calculate by hand are not correct.

The decision is to fail to reject the null given that p value is not ≤ 0.05.
The hypothesis
H0 = p = 0.63
H1 = p < 0.63
α = 0.05
sample proportion = 71/125 = 0.568
x = 71
n = 125
standard error of the proportion
√0.63(1-0.63)/125
= 0.0431
The null hypothesis follows a standard normal distribution. This is a left tailed test.
We are to reject the null if the p value is less than 0.05
p < 0.05
This probability test is a z probability test.
Z test = 0.568 - 0.63 / 0.0431
test statistic = -1.436
-Z0.05 =
Critical value = -1.645
p(z < -1.436)
p value = 0.0755
The decision would be to fail to reject the null hypothesis. The reason would be due to the fact that p value is greater than significance.
P-value is not ≤ α 0.05
0.0755 > 0.05
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