A survey of 340 randomly chosen US adults found that 60% of the 150 men and 50% of the 190 women ran a five-kilometer race during the past month. Do these data provide statistical evidence at the α = 0.05 level that women are less likely than men to run a five-kilometer race? Be sure to state the parameter, check conditions, perform calculations, and make conclusion(s). (10 points)

The p-value is less than the significance level, so we will reject the null hypothesis, and conclude that at 5% significance level, the proportion for women that ran 5 km is more than the proportion of male that ran the 5 km. Thus, women are more likely to run the 5 km race.

Step-by-step explanation:

We are given;

Number of males selected; n₁ = 150

Number of females selected; n₂ = 190

Proportion of the males; p₁ = 0.6

Proportion of females; p₂ = 0.5

Let's define the hypotheses;

Null hypothesis; H0: p₁ ≥ p₂

Alternative hypothesis; Ha: p₁ < p₂

Now, the z score formula for this is;

z = (p₁ - p₂)/√(p^(1 - p^))(1/n₁ + 1/n₂))

Where;

p^ = (p₁ + p₂)/2

p^ = (0.6 + 0.5)/2

p^ = 0.55

Thus;

z = (0.6 - 0.5)/√(0.55(1 - 0.55))((1/150) + (1/190))

z = 0.1/0.0543

z = 1.84

From online p-value from z-score calculator attached, using z = 1.84, significance level = 0.05, one tailed hypothesis, we have;

P-value = 0.033

The p-value is less than the significance level, so we will reject the null hypothesis, and conclude that at 5% significance level, the proportion for women that ran 5 km is more than the proportion of male that ran the 5 km. Thus, women are more likely to run the 5 km race.

Although if we check at 0.01 significance level, we will not have the same answer as the p-value will be greater.

ian gonna lie that is way to much