sing traditional methods, it takes 11.7 hours to receive a basic driving license. A new license training method using Computer Aided Instruction (CAI) has been proposed. A researcher used the technique with 23 students and observed that they had a mean of 11.3 hours with a standard deviation of 1.4. A level of significance of 0.1 will be used to determine if the technique performs differently than the traditional method. Assume the population distribution is approximately normal. Make the decision to reject or fail to reject the null hypothesis.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the CAI technique performs differently than the traditional method (P-value = 0.1844).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the CAI technique performs differently than the traditional method.

Then, the null and alternative hypothesis are:

The significance level is 0.1.

The sample has a size n=23.

The sample mean is M=11.3.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1.4.

The estimated standard error of the mean is computed using the formula:

Then, we can calculate the t-statistic as:

The degrees of freedom for this sample size are:

This test is a two-tailed test, with 22 degrees of freedom and t=-1.37, so the P-value for this test is calculated as (using a t-table):

As the P-value (0.1844) is bigger than the significance level (0.1), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the CAI technique performs differently than the traditional method.

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}