Question

make a claim based on the presented data in the table, then answer set questions about the claim.

AutoSave OFF OFF A SOCIET a Home Insert Draw Design Layout References Mailings Review View Tell me Share Comments Aalcedie Aarodie Ahchare Calibri Ligh... v 16 A A A A BIU XX' ADA AaBbCcDc AaBbcDdEt AaBondee Heading 1 Hercing 2 Heading 3 Paste Normal No Spacing Heading 4 Dictate Sensitivity Styles Pere EEE 2 == Food affordability Name of state Data value State 1 Arizona 7.8 State 2 Colorado 9.9 State 3 District of 6.2 Columbia State 4 Hawaii 11,2 State 5 Indiana 8.3 State 6 Kentucky 7.8 State 7 Maryland 6 State 8 Minnesota 9.7 State 9 Montana 6 State 10 New Hampshire 15.5 State 11 New York 10.2 State 12 Ohio 9.9 State 13 Pennsylvania 9.4 State 14 South Dakota 8.8 State 15 Utah 11.4 Hypothesis test: Make a claim- Based on the claim you made above, conduct a hypothesis test. Level of significance (choose from 5% or 1%): Null hypothesis. Ho: Alternative hypothesis - HH In words, clearly state what your random variable X represents: Which distribution you will use for the test, Normal or Student's t (ift- distribution, state the degrees of freedom): What is the test statistic? What type of test is used: two-tailed, right-tailed, left-tailed? What is the p-value? In one or two complete sentences, explain what the p-value means for this test: Page 1 of 1 167 wards English (United States Focus -- + 127%

Answer 1

Make a claim: The average food affordability of States is more than 9.

Level of significance:

g= 0,05

Null hypothesis:

Holi = 9

Alternate hypothesis: $H_{1}:\mu>9$

The random variable represents the sample mean of the food affordability.

X

Distribution: I shall use a t-distribution since the population variance is not knownwith n-1=14 degrees of freedom

Test statistic:

X- t= s/n

$t=\frac{9.2067-9}{2.4665/\sqrt{15}}$

$t=\frac{0.2067}{2.4665/3.8730}$

$t=\frac{0.2067}{0.6368}$

$t=0.3246$

Type of test: It is an one tailed test and is right tailed.

p-value: The p=0.3752

The p-value for this test means that around 38 in 100 times, we could have concluded that the population mean food affordability is more than 9 and in fact the true availability 9.

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For calculating the mean and variance, we make use of the table of sum and sum of squares.

State	Food affordability	x^2
1	7.8	60.84
2	9.9	98.01
3	6.2	38.44
4	11.2	125.44
5	8.3	68.89
6	7.8	60.84
7	6	36
8	9.7	94.09
9	6	36
10	15.5	240.25
11	10.2	104.04
12	9.9	98.01
13	9.4	88.36
14	8.8	77.44
15	11.4	129.96
Total	138.1	1356.61

Mean of x,

=1 TS ci η

$\bar{x}=\frac{138.1}{15}=9.2067$

Sample variance of x, $s^{2}=\frac{1}{n-1}\left [ \sum x_{i}^{2}-\frac{1}{n}\left ( \sum x_{i} \right )^{2} \right ]$

  $s^{2}=\frac{1}{15-1}\left [ 1356.61-\frac{1}{15}\left (138.1 \right )^{2} \right ]$

  $s^{2}=\frac{1}{14}\left [ 1356.61-\frac{19071.61}{15} \right ]$

  $s^{2}=\frac{1}{14}\left [ 1356.61-1271.4407 \right ]$

$s^{2}=\frac{1}{14}\left [ 85.1693 \right ]$

  $s^{2}=6.0835$

  $s=2.4665$