Mathematics, 16.11.2020 17:20, justiceemani12

It is known that a particular laboratory task takes the average person 2.5 seconds (the variance is not known). A researcher was interested in whether older people are slower (take longer) on this task. The researcher tested 30 randomly selected 80-year-olds. Their mean time was 2.7 seconds, with an estimated population standard deviation of 1.4 seconds. What should the researcher conclude?

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answered: IDONTHAVEABRAIN

Complete Question

It is known that a particular laboratory task takes the average person 2.5 seconds (the variance is not known). A researcher was interested in whether older people are slower (take longer) on this task. The researcher tested 30 randomly selected 80-year-olds. Their mean time was 2.7 seconds, with an estimated population standard deviation of 1.4 seconds. What should the researcher conclude? (Use the .05 significance level and use the five steps of hypothesis testing.)

The researchers conclusion is

There is no sufficient evidence to show that older people are slower at the laboratory task

Step-by-step explanation:

From the question we are told that

The mean time is $\mu = 2.5 \ seconds$

The sample size is n = 30

The sample mean is $\= x = 2.7 \ seconds$

The standard deviation is $\sigma = 1.4 \ seconds$

The level of significance is $\alpha = 0.05$

The null hypothesis is $H_o : \mu = 2.5$

The alternative hypothesis is $H_a : \mu 2.5$

Generally the test statistics is mathematically represented as

$z = \frac{\= x - \mu }{ \frac{\sigma}{\sqrt{n} } }$

=> $z = \frac{2.7 - 3 }{ \frac{1.4}{\sqrt{30} } }$

=> z =0.78

From the z table the area under the normal curve to the right corresponding to z =0.78 is

p-value = P(Z 0.78) = 0.2177

From the value obtained we see that the p-value > $\alpha$

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to show that the older people are slower at the laboratory task

answered: Guest

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

$|r| < 1$

which is equivalent to

$-1 < r < 1$

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

$s = \dfrac{a}{1 -r}$

where a is the first term.

$s = \dfrac{100}{1 - 1/2} = 200$

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}

$|x| < \frac 1 2 \sqrt[3]{2}$

answered: Guest

A. p = 2(l + w)p/2 - l = ww = p/2 - l b. while isolating the variable w in the equation, i first divided both sides by 2 and then i subtracted l from both sides. i hope this .