Krista has a quiz today. There are 4 questions with 4 options. Each question only has one correct answer. She wants to guess and

Answer:

The P-value ranges between 2.5% and 5% according to the t-table.

Step-by-step explanation:

A random sample of 16 students from a large university showed an average age of 25 years with a standard deviation of 2 years.

Let = true average age of all students at the university.

So, the Null Hypothesis, : 24 years {indicating the average age is less than or equal to 24 years}

Alternate Hypothesis, : > 24 years {indicating the average age is significantly greater than 24 years}

Here we employ the One-sample t-test statistics as the population's standard deviation is unknown;

                              T.S. = ~

where, = sample average age = 25 years

             s = sample standard deviation = 2 years

             n = sample size = 16

This gives us the test statistics = ~

                                     = 2

The value of the t-test statistics is 2.

Moreover, the P-value of the test-statistics can be found as follows;

P-value = P( > 2) = 0.034 {as per the t-table}

Thus, the P-value lies between 2.5% and 5% based on the t-table.

Answer: 1.2 < 1.7x

Step-by-step explanation:

Answer:

The ratio   corresponds to the tangent of ∠I.

Step-by-step explanation:

Let’s revisit the trigonometric ratios:

• sin Ф =  
• cos Ф =  
• tan Ф =  

For triangle HIJ

∵ m∠J = 90°

- The hypotenuse is the side opposite the right angle.

So, HI is the hypotenuse.

∵ HJ = 3 units

∵ IH = 5 units

- We’ll apply the Pythagorean Theorem to solve for HJ.

∵ (HJ)² + (IJ)² = (IH)²

∴ 3² + (IJ)² = 5²

∴ 9 + (IJ)² = 25

- Subtract 9 from both sides.

∴ (IJ)² = 16

- Taking the square root on both sides gives:

∴ IJ = 4 units

To determine the tangent of ∠I, identify the sides that are opposite and adjacent to it.

∵ HJ is opposite to ∠I

∵ IJ is adjacent to ∠I

- Utilizing the rule of tan above:

∴ tan(∠I) =

∴ tan(∠I) =

The ratio indicates the tangent of ∠I.

Response:

Detailed explanation:

For private institutions,

n = 20

Average, x1 = (43120 + 28190 + 34490 + 20893 + 42984 + 34750 + 44897 + 32198 + 18432 + 33981 + 29498 + 31980 + 22764 + 54190 + 37756 + 30129 + 33980 + 47909 + 32200 + 38120)/20 = 34623.05

Standard deviation = √(sum of (x - mean)²/n

Sum of (x - mean)² = (43120 - 34623.05)^2 + (28190 - 34623.05)^2 + (34490 - 34623.05)^2 + (20893 - 34623.05)^2 + (42984 - 34623.05)^2 + (34750 - 34623.05)^2 + (44897 - 34623.05)^2 + (32198 - 34623.05)^2 + (18432 - 34623.05)^2 + (33981 - 34623.05)^2 + (29498 - 34623.05)^2 + (31980 - 34623.05)^2 + (22764 - 34623.05)^2 + (54190 - 34623.05)^2 + (37756 - 34623.05)^2 + (30129 - 34623.05)^2 + (33980 - 34623.05)^2 + (47909 - 34623.05)^2 + (32200 - 34623.05)^2 + (38120 - 34623.05)^2 = 1527829234.95

Standard deviation = √(1527829234.95/20

s1 = 8740.22

For public institutions,

n = 20

Average, x2 = (25469 + 19450 + 18347 + 28560 + 32592 + 21871 + 24120 + 27450 + 29100 + 21870 + 22650 + 29143 + 25379 + 23450 + 23871 + 28745 + 30120 + 21190 + 21540 + 26346)/20 = 25063.15

Sum of (x - mean)² = (25469 - 25063.15)^2 + (19450 - 25063.15)^2 + (18347 - 25063.15)^2 + (28560 - 25063.15)^2 + (32592 - 25063.15)^2 + (21871 - 25063.15)^2 + (24120 - 25063.15)^2 + (27450 - 25063.15)^2 + (29100 - 25063.15)^2 + (21870 - 25063.15)^2 + (22650 - 25063.15)^2 + (29143 - 25063.15)^2 + (25379 - 25063.15)^2 + (23450 - 25063.15)^2 + (23871 - 25063.15)^2 + (28745 - 25063.15)^2 + (30120 - 25063.15)^2 + (21190 - 25063.15)^2 + (21540 - 25063.15)^2 + (26346 - 25063.15)^2 = 1527829234.95

Standard deviation = √(283738188.55/20

s2 = 3766.55

This involves two independent samples. Define μ1 as the mean out-of-state tuition for private institutions and μ2 as the mean out-of-state tuition for public institutions.

The random variable represents μ1 - μ2 = the difference between the mean out-of-state tuition for private vs. public institutions.

The hypothesis is established as follows. The correct choice is

-B. H0: μ1 = μ2; H1: μ1 > μ2

As the sample standard deviation is known, the test statistic is calculated using the t test formula:

(x1 - x2)/√(s1²/n1 + s2²/n2)

t = (34623.05 - 25063.15)/√(8740.22²/20 + 3766.55²/20)

t = 9559.9/2128.12528473889

t = 4.49

The method for finding degrees of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [8740.22²/20 + 3766.55²/20]²/[(1/20 - 1)(8740.22²/20)² + (1/20 - 1)(3766.55²/20)²] = 20511091253953.727/794331719568.7114

df = 26

The probability value is obtained from the t test calculator. It is

p value = 0.000065

Given that alpha, 0.01 > the p value, 0.000065, we will reject the null hypothesis. Hence, at a significance level of 1%, the mean out-of-state tuition for private institutions is statistically significantly greater than that of public institutions.