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3 years ago

7

If the mean I.Q. is 100 and the standard deviation of I.Q. scores is 15, then an I.Q. of 130 will have a z score (or standard sc

ore) of ___________________.

1 answer:

skad [1K]3 years ago

7 0

Answer: I.Q. of 130 will have a z score (or standard score) of 2.

Step-by-step explanation:

Let X be a random variable that represents the I.Q. score.

Formula for Z-score : $Z=\dfrac{X-mean}{standard \ deviation}$

As per given, mean I.Q. = 100 and the standard deviation =15

For X = 130,

$Z=\dfrac{130-100}{15}=\dfrac{30}{15}=2$

So, I.Q. of 130 will have a z score (or standard score) of 2.

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Alex_Xolod [135]

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Step-by-step explanation:

We are given that Forty-two subjects were randomly assigned to one of two treatment groups, 21 per group.

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<em>Let </em> $\mu_1$ <em> = population mean score for negative message</em>

<em /> $\mu_2$ <em> = population mean score for positive message</em>

SO, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1\geq \mu_2$ {means that a negative message results in a higher or equal mean score than positive message}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1$ {means that a negative message results in a lower mean score than positive message}

The test statistics that will be used here is <u>Two-sample t test statistics</u> as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n__1+_n__2-2$

where, $\bar X_1$ = sample mean score for negative message = 9.64

$\bar X_2$ = sample mean score for positive message = 15.84

$s_1$ = sample standard deviation for negative message = 3.43

$s_2$ = sample standard deviation for positive message = 8.65

$n_1$ = sample of subjects receiving the negative message = 21

$n_2$ = sample of subjects receiving the positive message = 21

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(21-1)\times 3.43^{2}+(21-1)\times 8.65^{2} }{21+21-2} }$ = 6.58

So, <u><em>the test statistics</em></u> = $\frac{(9.64-15.84)-(0)}{6.58 \times \sqrt{\frac{1}{21}+\frac{1}{21} } }$ ~ $t_4_0$

= -3.053

<em>Now at 0.05 significance level, the t table gives critical value of -1.684 at 40 degree of freedom for left-tailed test. Since our test statistics is less than the critical value of t as -3.053 < -1.684, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.</em>

Therefore, we conclude that a negative message results in a lower mean score than positive message.

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3 years ago

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<h2>Steps</h2>

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Next, I will be using the substitution method to remove one of the variables. To do this, firstly subtract both sides by y in the first equation:

$x=20-y\\0.5x+0.75y=11.50$

Next, since we know that x is equal to 20 - y, substitute it into the second equation and solve for y:

$0.5(20-y)+0.75y=11.50\\10-0.5y+0.75y=11.50\\10+0.25y=11.50\\0.25y=1.50\\y=6$

Now that we have the value of y, substitute it into either equation to solve for x:

$x+6=20\\x=14\\\\0.5x+0.75(6)=11.50\\0.5x+4.5=11.50\\0.5x=7\\x=14$

<h2>Answer</h2>

<u>In short, Angelo bough 14 apples and 6 bananas.</u>

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4 years ago