6.403 x 10 to the power of 8 in standard form

A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writin

Morgarella [4.7K]

Answer:

(a) We reject our null hypothesis.

(b) We fail to reject our null hypothesis.

(c) We fail to reject our null hypothesis.

Step-by-step explanation:

We are given that a certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hr.

A random sample of 18 pens is selected.

Let $\mu$ = true average writing lifetime under controlled conditions

So, Null Hypothesis, $H_0$ : $\mu \geq$ 10 hr {means that the true average writing lifetime under controlled conditions is at least 10 hr}

Alternate Hypothesis, $H_A$ : $\mu$ < 10 hr {means that the true average writing lifetime under controlled conditions is less than 10 hr}

The test statistics that is used here is one-sample t test statistics;

T.S. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean

s = sample standard deviation

n = sample size of pens = 18

n - 1 = degree of freedom = 18 -1 = 17

Now, the decision rule based on the critical value of t is given by;

If the value of test statistics is more than the critical value of t at 17 degree of freedom for left-tailed test, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the value of test statistics is less than the critical value of t at 17 degree of freedom for left-tailed test, then we will reject our null hypothesis as it will fall in the rejection region.

(a) Here, test statistics, t = -2.4 and level of significance is 0.05.

Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.

Here, clearly the value of test statistics is less than the critical value of t as -2.4 < -1.74, so we reject our null hypothesis.

(b) Here, test statistics, t = -1.83 and level of significance is 0.01.

Now, at 0.051 significance level, the t table gives critical value of -2.567 at 17 degree of freedom.

Here, clearly the value of test statistics is more than the critical value of t as -2.567 < -1.83, so we fail to reject our null hypothesis.

(c) Here, test statistics, t = 0.57 and level of significance is not given so we assume it to be 0.05.

Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.

Here, clearly the value of test statistics is more than the critical value of t as -1.74 < 0.57, so we fail to reject our null hypothesis.

8 0

3 years ago

Please answer <33 I would lava it :)

allsm [11]

Answer:

5 gallons

Step-by-step explanation:

3/4 + 6/8 + 20/8 + 1

6/8 + 6/8 + 20/8 + 8/8

40/8 = 5 gallons

3 0

3 years ago

The bike portion of an ironman competition is 112 miles. Your friend says that if you continue to bike at this pace, you will be

Roman55 [17]

We have no clue because you have not said what the speed/pace is. At a speed of 17.23 mph you will bike 112miles in 6.5 hours.

5 0

3 years ago

Suppose p" must approximate p with relative error at most 10-3 . Find the largest interval in which p* must lie for each value o

goblinko [34]

Answer:

$[p-|p|*10^{-3} \, , \, p+|p|* 10^-3]$

Step-by-step explanation

The relative error is the absolute error divided by the absolute value of p. for an approximation p*, the relative error is

r = |p*-p|/|p|

we want r to be at most 10⁻³, thus

|p*-p|/|p| ≤ 10⁻³

|p*-p| ≤ |p|* 10⁻³

therefore, p*-p should lie in the interval [ - |p| * 10⁻³ , |p| * 10⁻³ ], and as a consecuence, p* should be in the interval [p - |p| * 10⁻³ , p + |p| * 10⁻³ ]

8 0

3 years ago

Find the value of the lesser root of x2 + 12x + 35 = 0.

matrenka [14]

In quadratic equations,
a + b = -B
ab = C
where a and b are the roots, B is the second term, and C is the constant.
Substituting,
a + b = -12
ab = 35
The values of a and b from the equation is -7 and -5. Thus, the lesser root is -7.

8 0

3 years ago