The chi-square test statistic is χ2 = 18.68 and the P-value is between 0.0025 and 0.005. What conclusion should the researcher make? Use α = 0.05.

We don't know what the exact p-value is, but we are told that it's as large as 0.005 which is smaller than alpha = 0.05

Since the p-value is smaller than alpha, this means we reject the null hypothesis.

The way you can remember this is "if the p-value is low, then the null must go". By "low", I mean "smaller than alpha".

Recall that the p-value is the probability of observing that specific test statistic, or larger. So the chances of chi-squared being 18.68 or larger is a probability between 0.0025 and 0.005; there's a very small chance of this happening. The p-value is based entirely on the assumption that the null is correct. But if the null is correct, then the chances of landing on this are very small. We have a contradiction that basically leads to us concluding the null must not be the case. It's not 100% guaranteed of course, but it's fairly strong evidence.

In short, the p-value being smaller than alpha = 0.05 means we reject the null.

In order to accept the null, the p-value must be 0.05 or larger.

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.

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;