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Oduvanchick [21]

4 years ago

13

An automobile manufacturer is concerned about the possibility of a fault in the engine of a particular model of one of its cars.

This fault could potentially cause the engine to fail to start. The average number of cars per year that experience this fault is 4 cars. If the probability that 10 or more cars experience this fault in a year is at least 0.01, the manufacturer will issue a recall. Will the manufacturer issue a recall for this engine fault? Use a TI-83, TI-83 plus, or TI-84 calculator to find the probability.

1 answer:

Rama09 [41]4 years ago

6 0

The cars went to town was 4 x 10 is 40

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In Major League Baseball, the American League (AL) allows a designated hitter (DH) to bat in place of the pitcher, but in the Na

hammer [34]

Answer:

Step-by-step explanation:

From the given information:

Let:

$\mu_1$ represent the population mean no. for DH group.

$\mu_2$ represent the population mean no. for no DH group

$n_1$ represent the sample sizes of DH

$n_2$ represent the sample sizes of no DH

Sample size: $n_1 = n_2 = 20$

For both groups, the population standard deviation of runs scored = 2.54

i.e

$\sigma^2_1 = \sigma_2^2 = 2.54$

The null & alternative hypothesis;

$H_o : \mu_1 \le \mu_2 \\ \\ H_a: \mu_1 > \mu_2$

Level of significance:

The test statistics is:

$z = \dfrac{\bar x_1 - \bar x_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma_2^2}{n_2} }}$

where;

$\bar x_1$ = sample mean for DH group

$\implies \dfrac{1}{n_1} \sum \limit ^{n_1}_{i=1} x_1 \\ \\ = \dfrac{1}{20}(0+6+8+2+2+4+7+7+6+5+1+1+5+4+4+5+7+11+10+0) \\ \\ = \dfrac{92}{20}$

= 4.6

$\bar x_2$ = sample mean for no DH group

$\implies \dfrac{1}{n_2} \sum \limit ^{n_1}_{i=1} x_1 \\ \\ = \dfrac{1}{20}(3+6+2+4+0+5+7+6+1+8+12+4+6+3+4+0+5+2+1+4) \\ \\ = \dfrac{83}{20}$

= 4.15

Now:

$z = \dfrac{4.60- 4.15}{\sqrt{\dfrac{2.54^2}{20} + \dfrac{2.54^2}{20} }} \\ \\ = \dfrac{0.45}{\sqrt{0.32258 +0.32258 }} \\ \\ = \dfrac{0.45}{0.803219} \\ \\$

= 0.5602

Since the test is one-tailed by looking at that $H_a:$

The P-value = $P(Z > z)$

$\implies 1 - P(Z \le z) \\ \\ = 1 - P(Z \le 0.5602)$

Using Excel Function " =normdist(z)"; we have":

$= 1- 0.7123$

P-value = 0.2877

Decision rule: To reject $H_o$ , if $p-value < \alpha \ at \ 0.10$

Conclusion: SInce P = 0.2877 which is $> \ \alpha \ at \ 0.10.$

We fail to reject the $H_o$ and conclude that there is insufficient evidence to support the given claim that: $\text{more runs are scored in games for which DH is used.}$

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If Jodi paid $3.30 for a hamburger, how many toppings were on Jodi’s hamburger?

ioda

Answer:

1 topping...

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A basketball Ronald 300 m down a hill in 25 seconds. What is its average speed down the hill

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