Asample of 38 observations is selected from one population with a population standard deviation of 3.4. the sample mean is 100.5. a sample of 51 observations is selected from a second population with a population standard deviation of 5.8. the sample mean is 98.8. conduct the following test of hypothesis using the 0.05 significance level. h0 : μ1 = μ2 h1 : μ1 ≠ μ2
a. this is a -tailed test.
b. state the decision rule. (negative values should be indicated by a minus sign. round your answers to 2 decimal places.)
c. compute the value of the test statistic. (round your answer to 2 decimal places.
d. what is your decision regarding h0?
e. what is the p-value? (round your answer to 4 decimal places.)

a. two-tailed test

b. reject H0 if Z>1.96 or Z<-1.96

c. Z = 1.73

d. we have failed to reject the null hypothesis

e. P-value = 0.0418

Step-by-step explanation:

We will use a Z test to resolve this, an it will be a two-tailed test because the hypothesis statements are not indicating a specific direction for the significant difference (H0 : μ1 = μ2 ; H1 : μ1 ≠ μ2), this also means that the significanclevel will be divided between the both tails (2.5% en each tail for the rejection regions). See attached drawing for reference.

We need to find our critical value:

Zα/2 = Z(0.05/2) = 0.025

If we look for 0.025 in a Z table we will find that the critical value is 1.96 to the right, and by symmetry -1.96 to the left. So our decision rule will be to reject H0 if Z>1.96 or Z<-1.96

The Z test will be done using the next equation:

Z = (x⁻1 - x⁻2) - (μ1 - μ2) / √( σ²1/n1 + σ²2/n2

Because we are testing the null hypothesis we know that μ1 - μ2 must be zero if they are supposed to be equal (H0 : μ1 = μ2), so we calculate as follows:

Z = (100.5 - 98.8) - (0) / √( (3.4)²/38 + (5.8)²/51 = 1.7/0.9817 = 1.73

Z<1.96, therefore we have failed to reject the null hypothesis

The P-value for this test would be represented by the probability of Z being greater than 1.73, so we can look for it in any Z table, finding that its value is 0.0418, and because P-value>0.025 we again confirm that we don't have evidence statistically significant to reject the null hypothesis

283 square feet

step-by-step explanation:

good luck

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;