Graph on a number line. −6 over 7

Two 5-year girls, Alyse and Jocelyn, have been training to run a 1-mile race. Alyse's 1 mile time A is approximately Normally di

tatyana61 [14]

Answer:

1.7 × 10⁻⁴

Step-by-step explanation:

The question relates to a two sample z-test for the comparison between the means of the two samples

The null hypothesis is H₀: μ₁ ≤ μ₂

The alternative hypothesis is Hₐ: μ₁ > μ₂

$z=\dfrac{(\bar{x}_1-\bar{x}_2)-(\mu_{1}-\mu _{2} )}{\sqrt{\dfrac{\sigma_{1}^{2} }{n_{1}}-\dfrac{\sigma _{2}^{2}}{n_{2}}}}$

Where;

$\bar {x}_1$ = 13.5

$\bar {x}_2$ = 12

σ₁ = 2.5

σ₂ = 1.5

We set our α level at 0.05

Therefore, our critical z = ± 1.96

For n₁ = n₂ = 23, we have;

$z=\dfrac{(13.5-12)-(0)}{\sqrt{\dfrac{2.5^{2} }{23}-\dfrac{1.5^{2}}{23}}} = 3.5969$

We reject the null hypothesis at α = 0.05, as our z-value, 3.5969 is larger than the critical z, 1.96 or mathematically, since 3.5969 > 1.96

Therefore, there is enough statistical evidence to suggest that Alyse time is larger than Jocelyn in a 1 mile race on a randomly select day and the probability that Alyse has a larger time than Jocelyn is 0.99983

Therefore;

The probability that Alyse has a smaller time than Jocelyn is 1 - 0.99983 = 0.00017 = 1.7 × 10⁻⁴.

8 0

3 years ago

A machine that fills beverage cans is supposed to put 12 ounces of beverage in each can. The standard deviation of the amount in

erica [24]

Answer:

$\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54$

The degrees of freedom are:

$df=n-1=10-1=9$

Now we can calculate the p value using the alternative hypothesis:

$p_v =2*P(\chi^2 >13.54)=0.279$

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

Step-by-step explanation:

Assuming the following data:"12.14 12.05 12.27 11.89 12.06

12.14 12.05 12.38 11.92 12.14"

We can calculate the sample deviation with this formula:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$n=10$ represent the sample size

$\alpha=0.05$ represent the confidence level

$s^2 =0.0217$ represent the sample variance

$\sigma^2_0 =0.12^2= 0.0144$ represent the value to verify

Null and alternative hypothesis

We want to determine whether the standard deviation differs from 0.12 ounce, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 = 0.0144$

Alternative hypothesis: $\sigma^2 \neq 0.0144$

The statistic can be calculated like this;

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

$\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54$

The degrees of freedom are:

$df=n-1=10-1=9$

Now we can calculate the p value using the alternative hypothesis:

$p_v =2*P(\chi^2 >13.54)=0.279$

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

7 0

3 years ago

Find the common difference of the arithmetic sequence, the next three terms, the equation, and the term given for 8, 10, 12, 14

fredd [130]

Answer:

see explanation

Step-by-step explanation:

In an arithmetic sequence the common difference d is

d = a₂ - a₁ = 10 - 8 = 2

To obtain the next term in the sequence add d to the previous term, that is

a₅ = 14 + 2 = 16

a₆ = 16 + 2 = 18

a₇ = 18 + 2 = 20

The next 3 terms in the sequence are 16, 18, 20

The n th term equation for an arithmetic sequence is

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 8 and d = 2, thus

$a_{n}$ = 8 + 2(n - 1) = 8 + 2n - 2 = 2n + 6

5 0

3 years ago

Find the common factors of the given terms: 2y, 22xy. Ok

Lesechka [4]

Answer:

2y

Step-by-step explanation:

this is common factor of both the terms

because 2y divides both term

hope it helps

5 0

3 years ago

Please help! Need em doneee

kogti [31]

Letter C

Use the Distance formula
sq rt [ (x2 - x1)^2 + (y2 - y1)^2 ]

sq rt [ (10-5)^2 + (2 - 4)^2 ]
sq rt [ (5)^2 + (-2)^2 ]
sq rt ( 25 + 4)
Square root of 29

6 0

1 year ago

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