Question

비 Alther ; a zoo cost who s dles Anna humming ird C ypt anna Kc arance: Hummingbirds L Lana, w. Alther Suppo e that n ล rcmote part o nc Grand Canyon, a random sampe af six C thCSC bir s was caught. wcIRhed, and cloa the w qhts in Rra s wcrc as follows. J.2. 8 4.2 4. 1 The sample m oan is x 3.75 0 ams. Let x be a r da iniable re ting ei ts at hummin birds in this pอ 0t the and Canyon we ass me thatx has a normal distribution and σ wcIght lsu-4. 3l) arams. Da ta, data Indicate that the mean wei nt cf thec hlrds In this part ofthc Grand Canyon ls less tnan 4.30 grams? uso:区一0.1a. 0 64 ฎ อ ท. Supo se it is knaw tt at tor e population ot all Anna's hummi ฏ ids, the mean Whar tha level af signlficance(Fnter a numher.) Slate the null and ullenele hypotheses Will you use egt-ailed, or two teiled lest? O HC: -4.3 g; H.: " > 4.3 g; right tailed (b) what sermping distribution will you use? Lxplain the rationale fer your chole of samoling distribution. C he student's t, s nce we assurne that x has a normal dstribution with known σ. Ο The standam norna, slnca we a sume trat x has nomai distribution with icnawn T. (c) rind (o' esLimeue} 1tre A-value. (С"Le' nuntve. Round You' answer lo four deurt'el places.) Sketch the sampling dIntribution and thaw the area enme;ponding to th。/yalue. (Salect the cnmect Groh.) 321 -3-2-1 0 ALth1เส-0. 10 level, we telecl lhe tlul hypothesis d"ป uonclude uve dstd ere sleusucally signficant. OAt tha . eel, w raect the nul hypothosks and conclude the dataa not statictically slanificant. O Atthc α-0.10 level, wc fall to rcect the null hypothesis and conclude the data are statistically significant. O At tha α-0.10 level, we tail to relert the null hrpathesis and conclude the data ar. not statistically sigriiant O There is suricienLeviderue ㅂLLhe 0.10 level w wno lude Ilie, tur"""'עו birus in Ihe Grand Ceriyon negl' less uten 4.30 g'd'rs

Answer 1

Solution

Let X = weight of Hummingbird in grams.

We are given X ~ N(µ, σ²), where σ is given to be 0.64 and µ is under test. ............................................................ (1)

Also given are: n = 6, Xbar = 3.75, Significance level, α = 0.10.................................................................................(2)

Part (a)

Sub-part (i)

Level of significance = 0.10 or 10% [vide (2)] Answer 1

Sub-part (ii)

This is a one-sided (< type i.e., left tail) test Answer 2

[‘ .........less than 4.30 gram ..... in the last sentence of the question is suggestive.]

Sub-part (iii)

Hypotheses:

Null H₀: µ = µ₀ = 4.30   Vs Alternative H_A: µ < 4.30 First Option Answer 3

Part (b)

Sub-part (i)

Vide (1), sample average Xbar is Normally distributed. Fourth Option Answer 4

Sub-part (ii)

Test statistic:

Z = (√n)(Xbar - µ₀)/σ, where n = sample size; Xbar = sample average; σ = known population standard deviation.

= - 2.11 [vide (1) and (2)] Answer 5

Part (c)

Sub-part (i)

Distribution and p-value

Under H₀, Z ~ N(0, 1) and hence p-value = P(Z < Zcal)

Using Excel Function, Statistical NORMSDIST, p-value is found to be: 0.1764 Answer 6

Sub-part (ii)

The p-value is correctly depicted in the second graph Answer 7

Part (d) and (e)

Sub-part (i)

Decision:

Since p-value < α. H₀ is rejected. Answer 7

Sub-part (ii)

Conclusion:

There is sufficient evidence to support the claim that

the mean weight of the Hummingbird population is less than 4.3 grams. Answer 8

First Option

DONE