Type II error is Fail to reject the claim that the proportion of people who write with their left hand is 0.29 when the proportion is different from 0.29.

Step-by-step explanation:

We are given the following hypothesis below;

Let p = proportion of people who write with their left hand

So, Null Hypothesis, $H_0$ : p = 0.22 {means that the proportion of people who write with their left hand is equal to 0.22}

Alternate Hypothesis, $H_A$ : p $\neq$ 0.22 {means that the proportion of people who write with their left hand is different from 0.22}

Now, Type I error states that we conclude that the null hypothesis is rejected when in fact the null hypothesis was actually true. Or in other words, it is the probability of rejecting a true hypothesis.

So, in our question; Type I error is to Reject the claim that the proportion of people who write with their left hand is 0.29 when the proportion is actually 0.29.

Type II error states that we conclude that the null hypothesis is accepted when in fact the null hypothesis was actually false. Or in other words, it is the probability of accepting a false hypothesis.

So, in our question; Type II error is Fail to reject the claim that the proportion of people who write with their left hand is 0.29 when the proportion is different from 0.29.

4 0

3 years ago

The graph of an equation with a negative discriminant always has which characteristic?

jasenka [17]

Answer:

a posative

Step-by-step explanation:

3 0

3 years ago

A fair die is tossed 5 times. Let <img src="https://tex.z-dn.net/?f=%5Cdfrac%7Bm%7D%7Bn%7D" id="TexFormula1" title="\dfrac{m}{n}

Ray Of Light [21]

Answer:

m = 1

Step-by-step explanation:

We can suppose that the number we are looking for is for example 5.

(we can do so because the probability is the same for each number - it'sna fair dice)

For the first toss the probability we have 5 is 1/6 (we have 6 numbers on the dice and number 5 is just one of the possible 6 outcomes).

For the second toss the probability we have 5 is again 1/6.

For the rest of 3 tosses we don'tcare what number we will get( we have our two consecutive 5s), so all of the outcomes for the rest of 3 tosses are good for us (probability is 6/6 = 1)

Threfore, the probability to get two consecutive 5s is 1/6 * 1/6 * 1 * 1 * 1 = 1/36.

We can see that m = 1.

4 0

3 years ago