Suppose that 40 students have done a test; here are the results obtained. the grade mean of the entire school population is 90. we would like to check if there is a statistically significant difference (with a confidence level of 90%) between the means of the sample and the population, assuming that the population variance is known and equal to 15. (65, 78, 88, 55, 48, 95, 66, 57, 79, 81 75, 58, 18, 35, 58, 96, 61, 52, 39, 44 61, 38, 68, 65, 88, 75, 46, 97, 72, 82 62, 78, 87, 50, 55, 99, 65, 53, 70, 84) use a one sample z-test, to solve the problem, and discuss if you accept the null hypothesis and why. z. test(x = ? , mu = ? , sd = ? )

Z-test is a statistical test used to check weather two means are significant or not for unknown variance and large sample size.

Z = (M - μ) ÷ √(σ² / n)

where, M = Sample Mean = 66.075

μ = Population Mean = 90

σ² = Population Variance = 15

Now, Calculating the value of Z-test:

Z = (66.075 - 90) ÷ √(225 ÷ 40)

Z = -23.925 ÷ 2.37171

Z = -39.06936

The value of Z is -39.06936.

The value of p is < .00001 at 90% confidence level.

If the p-value is less than value of α, We accept the null- hypothesis otherwise reject.

The result is significant at p < .10.

Thus, we accept the null hypothesis.

24

step-by-step explanation:

if it takes 8 pounds to make 12 divide 12 by 8 and you get 1.5, 1.5 x 16 = 24