At midnight, the temperature was -8oF. At noon, the temperature was 23oF. Which expression represents the increase in temperatur
2 answers:
Answer:
Step-by-step explanation:
We have been given that at midnight, the temperature was . At noon, the temperature was
.
Since our temperature has increased to 23 degree Fahrenheit from -8 degree Fahrenheit, we can represent this increase in temperature as:
Therefore, the expression represents the increase in temperature.
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121 is big enough to assume normality and not worry about the t distribution. By the 68-95-99.7 rule a 95% confidence interval includes plus or minus two standard deviations. So 95% of the cars will be in the mph range
The question is a bit vague, but it seems we're being asked for the 95% confidence interval on the average of 121 cars. The 121 is a hint of course.
The standard deviation of the average is in general the standard deviation of the individual samples divided by the square root of n:
So repeating our experiment of taking the average 121 cars over and over, we expect 95% of the averages to be in the mph range
That's probably the answer they're looking for.
Answer:
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.
