Auniversity is interested in promoting graduates of its honors program by establishing that the mean gpa of these graduates exceeds 3.50. a sample of 36 honors students is taken and is found to have a mean gpa equal to 3.60. the population standard deviation is assumed to equal 0.40. to establish whether the mean gpa exceeds 3.50, the appropriate hypotheses

H0:

H1:

Step-by-step explanation:

1) Data given and concepts involved

n=36 the sample size

sample mean obtained

The null hypothesis "Is what we attempt to find evidence against in our hypothesis test". And is denoted by H0

The alternative hypothesis "Is what we are attempting to demonstrate". and is denoted by H1.

In our case since we want to see if then that would be the alternative hypothesis, and the null hypothesis it's always the complement for the alternative hypothesis, so on this case

The distribution for is given by:

2) Test the hypothesis

In order to test the hypothesis we can use the following statistic:

Replacing the values given we have:

We know that and we can calculate a critical value, but we don't know the significance level for the hypothesis, so on this case we can't proceed.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}