In January 1995, each student in a random sample of 150 physics majors at a large university was asked in how many lab classes he or she was enrolled. The results indicated a mean of 1.74 lab classes and a standard deviation of 1.49. To determine whether the distribution changed over the past 20 years, a similar survey was conducted in January 2015 by selecting a random sample of 150 physics majors. The results indicated a sample mean of 1.87 and a standard deviation of 1.57. Do the data provide evidence that the mean number of lab classes taken by physics majors in January 1995 is different from the mean number of lab classes taken in 2015? Perform an appropriate statistical test using α = 0.10. (10 points)

The data provide evidence that the mean number of lab classes taken by physics majors in January 1995 is not different from the mean number of lab classes taken in 2015.

Step-by-step explanation:

A t-test for difference between means can be used to determine whether the mean number of lab classes taken by physics majors in January 1995 is different from the mean number of lab classes taken in 2015.

The hypothesis is:

H₀: There is no difference between the two population means, i.e. μ₁ = μ₂.

Hₐ: There is a significant difference between the two population means, i.e. μ₁ ≠ μ₂.

The survey was conducted in a similar pattern. Assume that the two population standard deviations are same.

The information provided is:

The test statistic is:

Here s is the pooled standard deviation.

Compute the value of pooled standard deviation as follows:

Compute the test statistic value as follows:

The test statistic value is -0.735.

The decision rule is:

If the p - value of the test is less than the significance level, α = 0.10 then the null hypothesis will be rejected and vice-versa.

Compute the p-value as follows:

The p-value = 0.463 > α = 0.10.

As the p-value is more than the significance level the null hypothesis will not be rejected at 10% level of significance.

Conclusion:

The data provide evidence that the mean number of lab classes taken by physics majors in January 1995 is not different from the mean number of lab classes taken in 2015.

a) $ 9.50

b) 17

c) y-intercept (0, 12). this is the initial amount of money that joe had prior to playing any game.

d) x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

e) slope is -0.5. joe's balance will reduce by 0.5 for every game that he plays.

f)

domain; [0,24]

range; [0,12]

g) x + 2y = 24

step-by-step explanation:

a)

if joe plays 5 games, this will imply that x = 5. substitute x = 5 in the equation and solve for y;

y = -.50x + 12

y = -0.50(5) + 12

y = 9.50

b)

if joe leaves the arcade with $3.50, this will mean that y = 3.50. plug in y = 3.50 in the given equation and solve for x;

y = -.50x + 12

3.50 = -0.50x + 12

0.50x = 12 - 3.50

0.5x = 8.5

x = 17

c)

the function given is   y = -.50x + 12.

the y-intercept is the point where the graph of this function will intersect the y-axis. at this point, the value of x is usually 0. plug in x = 0 in the given equation and determine y;

y = -0.50(0) + 12

y = 12

the y-intercept is thus (0, 12). this is the initial amount of money that joe had prior to playing any game.

d)

the x-intercept refers to the point where the graph of a function intersects with the x-axis. at this point, the value of y is 0. substitute y = 0 in the given equation and solve for x;

0 =   -.50x + 12

0.5x = 12

x = 24

x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

e)

the slope of the given equation is -0.5. the equation has been given in slope-intercept form. thus the slope is simply the coefficient of x.

joe's balance will reduce by 0.5 for every game that he plays. this is the interpretation of the slope.

f)

domain is the set of x values for which the function is real and defined. the domain will thus be; [0,24]. this will restrict the balance to only assume positive values.

on the other hand, the range is the set of y values for which a function is real and defined. the range in our case will be [0,12]. joe can have between $0 and $12 balance.

g)

the standard form of the equation of a straight line is;

ax + by = c

where a, b, and c are integers.

our equation is given as;

y = -.50x + 12

re-writing the equation we have;

0.5x + y = 12

divide all through by 0.5;

x + 2y = 24

answer:

c 60% on edg