Can someone help me with eatheR G AND H, I AND J, OR. K AND L plzzzz two of any of your choice :)

5) In a certain supermarket, a sample of 60 customers who used a self-service checkout lane averaged 5.2 minutes of checkout tim

Ne4ueva [31]

Answer:

$\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643$

$S_p=2.940$

$t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751$

$df=60+72-2=130$

$p_v =P(t_{130}$

Assuming a significance level of $\alpha=0.05$ we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

Step-by-step explanation:

Data given

Our notation on this case :

$n_1 =60$ represent the sample size for people who used a self service

$n_2 =72$ represent the sample size for people who used a cashier

$\bar X_1 =5.2$ represent the sample mean for people who used a self service

$\bar X_2 =6.1$ represent the sample mean people who used a cashier

$s_1=3.1$ represent the sample standard deviation for people who used a self service

$s_2=2.8$ represent the sample standard deviation for people who used a cashier

Assumptions

When we have two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

The statistic is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

And t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

System of hypothesis

Null hypothesis: $\mu_1 \geq \mu_2$

Alternative hypothesis: $\mu_1 < \mu_2$

This system is equivalent to:

Null hypothesis: $\mu_1 - \mu_2 \geq 0$

Alternative hypothesis: $\mu_1 -\mu_2 < 0$

We can find the pooled variance:

$\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643$

And the deviation would be just the square root of the variance:

$S_p=2.940$

The statistic is given by:

$t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751$

The degrees of freedom are given by:

$df=60+72-2=130$

And now we can calculate the p value with:

$p_v =P(t_{130}$

Assuming a significance level of $\alpha=0.05$ we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

5 0

3 years ago

Angelo earns $2,080 each month. His total deductions are 30% of his pay. How much is deducted from his pay each month?

Fofino [41]

C. $693.33. Just round. It helps. Round 2,080 to 2,100. Then whats 30% of 2,100? It's easy just think of 21 divided by 3, which equals 7. So, 700 and the closest answers to that is $693.33. That's just the easy way instead of doing the math out. And it is also common sense that B or D wouldn't be it. So your left with C or A. :)

7 0

3 years ago

Read 2 more answers

Suppose j varies jointly with g and v, and j=1 when g=4 and v=5. what is j when g=10 and v=9?

Tems11 [23]

J $\alpha gv$
j=agv where a is a constant of proportionality.
j=1 when g=4 and v=5
1=a*4*5
1=20a
a=1/20
a= 0.05
j=0.05gv
When g=10 and v=9,
j=0.05*10*9
j=0.5*9
j=4.5

4 0

3 years ago

The results of a survey asking 500 people about their favorite ice cream flavor are shown in the table.

brilliants [131]

chocolate: experimental 100 theoretical 165

vanilla: experimental 100 theoretical 135

Mint chip: experimental 100 theoretical 100

banana:experimental 100 theoretical 30

<em>BRAINLYEST PLZ PLZ PLEASE </em>

7 0

3 years ago

A manufacturer of matches randomly and independently puts 23 matches in each box of matches produced. The company knows that one

d1i1m1o1n [39]

Answer:

0.9855 or 98.55%.

Step-by-step explanation:

The probability of each individual match being flawed is p = 0.008. The probability that a matchbox will have one or fewer matches with a flaw is the same as the probability of a matchbox having exactly one or exactly zero matches with a flaw:

$P(X\leq 1)=P(X=0)+P(X=1)\\P(X\leq 1)=(1-p)^{23}+23*(1-p)^{23-1}*p\\P(X\leq 1)=(1-0.008)^{23}+23*(1-0.008)^{23-1}*0.008\\P(X\leq 1)=0.8313+0.1542\\P(X\leq 1)=0.9855$

The probability that a matchbox will have one or fewer matches with a flaw is 0.9855 or 98.55%.

8 0

3 years ago